What are the odds of getting the following hands in euchre?
Solutions (TLDR)

The probabilities of a single player getting one of the above hands, from a single deal, are:


Scenario 1: 0.09%
Scenario 2: 1.54%
Scenario 3: 0.05%


Below I have calculated the probabilities and also simulated them to check the calculations.

I have also simulated the scenarios to find the probabilities of any player getting one of the above hands, from a single deal.  I think this more closely matches the observations described in the question.  These probabilities are:


Scenario 1 for any player: 0.38%
Scenario 2 for any player: 5.98%
Scenario 3 for any player: 0.19%


Introduction

A standard euchre deck contains the 9, 10, J, Q, K and A in each of the four suits, making a total of 24 cards.  Each of the four players are dealt a hand of five cards, and there are four cards left over.

We will split the deck, D, into the four Jacks J, and the other twenty non-Jack cards N.  We refer to the player's hand as H, and the table cards as T.

We use the notation nCk to refer to the Combinations function; the number of ways of choosing k elements from a set of size n.

We refer to the probability of an event using the notation P(Event).  We will refer to the conditional probability of event 2 happening, given that event 1 happens, with the notation P(Event 2 | Event 1).  In several places we do not explicitly state that an event is conditional on another event, but hopefully the context makes it obvious when this is the case.

Scenario 1: What are the odds that you get dealt both Jacks of the same color, plus three other cards all of the same suit in the same color? (i.e. Jack of Hearts, Jack of Diamonds, plus Ace, Queen and Ten of Diamonds).

P(Scenario1) = P(H contains exactly 2 Jacks) * P(The Jacks are the same colour) 
               * P(3 other cards are same suit) * P(The suit is the same colour as the Jacks)
P(H contains exactly 2 Jacks) = (Ways to pick 2 from J) * (Ways to pick 3 from N) 
                                / (Ways to pick 5 from D)
                              = (4C2)(20C3) / (24C5)
                              = 6*1140 / 42504
P(The Jacks are the same colour) = 1/3
P(3 other cards are same suit) = P(The second card is the same suit as the first)
                                 * P(The third is the same suit as the other two)
                               = 4/19 * 3/18
P(The suit is the same colour as the Jacks) = 1/2
∴ P(Scenario1) = 6*1140 / 42504 * 1/3 * 4/19 * 3/18 * 1/2
               = 5/5313
               ~= 0.09%


Scenario 3: What are the odds you get dealt all 4 Jacks?

This calculation is slightly more straightforward than for Scenario 1, as we no longer need to worry about suits.

P(Scenario3) = P(H contains exactly 4 Jacks)
             = (4C4)(20C1)/(24C5)
             = 1*20 / 42504
             = 5/10626
             ~= 0.05%


Scenario 3 is exactly half as likely as scenario 1.

Scenario 2: What are the odds you get dealt four of the cards asked in Scenario 1, and the fifth card you need is the face-up card in the pile of 4 remaining cards?

This is harder to compute.  First we split the calculation down by distribution of H.  I've written d(H)=(3,2) to mean that H contains 3 Jacks and 2 non-Jacks. Note that it is only possible to get Scenario 2 if H contains 1,2 or 3 Jacks.  We refer to "Scenario 2" as S2 for brevity.

P(S2) = P(d(H)=(3,2))P(S2 | d(H)=(3,2)) + P(d(H)=(2,3))P(S2 | d(H)=(2,3))
        + P(d(H)=(1,4))P(S2 | d(H)=(1,4))
P(d(H)=(j,n)) = (4Cj)(20Cn) / (24C5)


Scenario 2, Part 1: Three Jacks in the hand

Consider the case where H contains 3 Jacks and 2 non-Jacks.

P(S2 | d(H)=(3,2)) = P(2 of the Jacks are same colour) * P(2 non-Jacks are same suit)
                     * P(Suit of non-Jacks is same colour as 2 Jacks)
                     * P(T contains at least one non-Jack of same suit)
P(2 of the Jacks are same colour) = 1
P(2 non-Jacks are same suit) = 4/19
P(Suit of non-Jacks is same colour as 2 Jacks) = 1/2
P(T contains at least one non-Jack of same suit) = 1 - P(T contains 0 non-Jacks of same suit)
                                                 = 1 - 16/19 * 15/18 * 14/17 * 13/16
                                                 = 514/969
∴ P(S2 | d(H)=(3,2)) = 1 * 4/19 * 1/2 * 514/969 = 1028/18411


Scenario 2, Part 2: Two Jacks in the hand

Next consider the case where H contains 2 Jacks and 3 non-Jacks.  We treat the cases where the Jacks are the same colour separately from the cases where they are different.

P(S2 | d(H)=(2,3)) = P(Jacks are same colour)P(S2 | d(H)=(2,3) and Jacks are same colour)
                     + P(Jacks not same colour)P(S2 | d(H)=(2,3) and Jacks not same colour)
                   = 1/3 P(S2 | d(H)=(2,3) and Jacks are same colour)
                     + 2/3 P(S2 | d(H)=(2,3) and Jacks not same colour)


When the Jacks are the same colour, we need exactly two of the three non-Jacks to share a suit, and that suit to be the same colour as the Jacks.

P(S2 | d(H)=(2,3) and Jacks are same colour) = P(Exactly 2 non-Jacks share a suit)
  * P(Suit is same colour as Jacks) * P(T contains at least one non-Jack of same suit)
  = (3*(4*15)/(18*19)) * (1/2) * (514/969)
  = 2570/18411


When the Jacks are different suits, we need all three non-Jacks to share a suit.

P(S2 | d(H)=(2,3) and Jacks not same colour)
 = P(All 3 non-Jacks share a suit) * P(T contains other Jack of same colour)
 = (4*3)/(19*18) * 4/19
 = 8/1083


Putting these two cases together we get:

P(S2 | d(H)=(2,3)) = 1/3 * 2570/18411 + 2/3 * 8/1083 = 2842/55233


Scenario 2, Part 3: One Jack in the hand

The final distribution of H to consider is 1 Jack and 4 non-Jacks.

P(S2 | d(H)=(1,4)) = P(At least 3 non-Jacks share a suit) * P(The Jack is the same colour)
                     * P(T contains other Jack of same colour)
P(At least 3 non-Jacks share a suit) = (4*4*3*15 + 4*3*2) / (19*18*17) = 124/969
P(The Jack is the same colour) = 1/2
P(T contains other Jack of same colour) = 4/19
∴ P(S2 | d(H)=(1,4)) = 124/969 * 1/2 * 4/19 = 248/18411


Scenario 2: Conclusion

So putting all of that together.

P(S2) = (4C3)(20C2)/(24C5) * 1028/18411 + (4C2)(20C3)/(24C5) * 2842/55233
        + (4C1)(20C4)/(24C5) * 248/18411
      = 4*190/42504 * 1028/18411 + 6*1140/42504 * 2842/55233 + 4*4845/42504 * 248/18411
      = 4670/302841
      ~= 1.54%


Simulation

To check my calculations I have written a Python script to evaluate ten million deals, to see how frequently each of the scenarios occurs.  The script is here.

I also used it to compute the probability of any of the four players satisfying each of the three scenarios for a given deal.  I thought this would be interesting, as this is what was observed in the question, although not actually the probability being asked for.

Probability for a single player   | Probability across all four players
Scenario1 | Scenario2 | Scenario3 | Scenario1 | Scenario2 | Scenario3
 0.0972%  |  1.5319%  |  0.05  %  |  0.3703%  |  5.9367%  |  0.1748%
 0.0927%  |  1.519 %  |  0.0442%  |  0.3716%  |  5.9698%  |  0.1914%
 0.0911%  |  1.5607%  |  0.0466%  |  0.3896%  |  5.9523%  |  0.187 %
 0.096 %  |  1.5207%  |  0.0442%  |  0.3753%  |  5.9583%  |  0.1853%
 0.0946%  |  1.5594%  |  0.0457%  |  0.3711%  |  5.9709%  |  0.1844%
 0.1004%  |  1.561 %  |  0.0466%  |  0.3824%  |  6.0065%  |  0.1916%
 0.0964%  |  1.5367%  |  0.0476%  |  0.3776%  |  5.9651%  |  0.1882%
 0.098 %  |  1.563 %  |  0.0452%  |  0.3775%  |  6.0131%  |  0.188 %
 0.0942%  |  1.5385%  |  0.0469%  |  0.3849%  |  6.0205%  |  0.1852%
 0.0983%  |  1.535 %  |  0.0418%  |  0.3641%  |  5.9913%  |  0.1931%
-----------------------------------------------------------------------
 0.0959%  |  1.5426%  |  0.0459%  |  0.3764%  |  5.9785%  |  0.1869%


Happily the simulation backs up the calculations.