How do you calculate hand probabilities for 10 cards per hand (8 suits); e.g.: Cripple Mr Onion?
I find hypergeometric series nice for computing, but poor for explaining.  For teaching purposes, I much prefer expanding these out using binomial coefficients.  Some books, and Wolfram Alpha call this "choose."

The key here is to realize that order does not matter.  The total number of possible hands is choose(104, 10) = 26 trillion.

Now, to count how many hands have 2 aces, you need to figure out how many ways there are to choose 2 aces, and multiply that by the number of ways to get 8 cards with no aces.  Writing it out in detail we have:

                    # of hands w 2 aces    
P(2 aces in hand) = -------------------
                        # of hands

     # ways to choose 2 aces * # ways to choose non aces
  = ----------------------------------------------------
                      choose(104,10)

     choose(8,2) * choose(96,8)
  = ----------------------------  = 0.14
           choose(104,10)


Here's the Wolfram Alpha expression for all the accuracy you could ever want.

Your second question, how to find {4,5,7} in one of two suits is ambiguous.  Do all three have to be in the same suit?  Or different suits?  There's another issue.  Let's call the suits ABCDEFGHIJ, and say you want the 4,5,7 to be in suits A or B.  Do you care if there is a 4 in suit F?  

P(4,5,7 in either of 2 suits, don't care about other suits) = 

     # ways to choose {4,5,7} * number of ways to choose other cards
  = ----------------------------------------------------
                      choose(104,10)

         2*2*2*choose(98,7)
  = ----------------------------  = 0.004
           choose(104,10)


Or, the related:
    P(4,5,7 in either of 2 suits, no other 4,5,7 in hand) = 

     # ways to choose {4,5,7} * number of ways to choose other cards
  = ----------------------------------------------------
                      choose(104,10)

         2*2*2*choose(80,7)
  = ----------------------------  = 0.001
           choose(104,10)


You might reasonably ask, why are these numbers so much lower than for aces?  This is because you are specifying three cards in the hand, and the aces are only specifying two cards in the hand.

Keep in mind, that in most games, this isn't the full story.  For instance, the way I've computed things, I've included a pair of aces and three kings.  In poker, that's a full house, and so needs to be removed from this probability.  If that's what you want, then things get more complicated.