Simple Task Solved Thrice
J, 16 bytes

Function 1, 5 bytes

p:^i.


Function 2, 6 bytes

+p:^i.


Function 3, 5 bytes

&gt;:^i.


How it works

Function 1

p:^i.     Right argument: y

   i.     Compute (0 ... y-1).
p:        Compute P, the prime at index y (zero-indexed).
  ^       Return all powers P^e, where e belongs to (0 ... y-1).


Since P is prime and P > y, y cannot divide Pe.

Function 2

+p:^i.    Right argument: y

 p:^i.    As before.
+         Add y to all results.


If y divided Pe + y, it would also divide Pe + y - y = Pe.

Function 3

&gt;:^i.     Right argument: y

   i.     Compute (0 ... y-1).
&gt;:        Compute y+1.
  ^       Return all powers (y+1)^e, where e belongs to (0 ... y-1).


If y divided (y+1)e some prime factor Q of y would have to divide (y+1)e.

But then, Q would divide both y and y+1 and, therefore, y + 1 - y = 1.