Part (a) is pretty straightforward. Take an ace out, and you have 51 cards, 3 of which are aces, so that's 1/17, and then the question is basically just what's the probability of an ace being in 12 of those , so 1/17 * 12/17.
It's not quite like that. If we work with unordered hands then there are (52 choose 13) hands in total, of which (48 choose 13) have no aces. So (52 choose 13) - (48 choose 13) hands contain at least one ace. All we know is that our hand is one of these.
Of these, 4 x (48 choose 12) hands contain exactly one ace. So,
courtesy Wolfram Alpha (
EDIT:
try this instead), the probability of at least two aces given that we have at least one ace is about
0.5071 0.3696.
For (b), the number of hands containing the ace of spades is (51 choose 12). Of these, (48 choose 12) contain no additional aces. So, again
courtesy Wolfram Alpha, the probability of at least two aces given the ace of spades is about 0.5612.
Your analysis for (a) is closer to the correct argument for (b). You are indeed drawing 12 cards from 51, of which 3 are aces, so the average number of aces you expect to draw is 12/17. But sometimes you'll get more than one ace, so the probability of
at least one ace has to be a bit lower to compensate.
