a) Assuming a standard deck of 52 cards. By definition of conditional probability:
P(3 spades | ace of spades) = P(3 spades AND ace of spades) / P(ace of spades)
We draw 3 cards, so there are (52 choose 3) = 22,100 total possible hands.
There are (1 choose 1)*(51 choose 2) = 1275 possible hands with the ace of spades.
There are (13 choose 3) = 286 possible hands containing 3 spades, and (1 choose 1)*(12 choose 2) = 66 of these contain the ace of spades.
So we have
P(3 spades AND ace of spades) = 66/22,100
P(ace of spades = 1275/22,100
and so
P(3 spades | ace of spades) = 66/1275 = 0.0518
b) By definition:
P(3 spades | at least 1 spade) = P(3 spades AND at least 1 spade) / P(at least 1 spade)
If you draw 3 spades, you already meet the requirement of having at least 1 spade, so
P(3 spades AND at least 1 spade) = P(3 spades)
and we already know there are 286 possible such hands.
The event of getting at least 1 spades is complementary to the event of not getting any spades, meaning
P(at least 1 spade) = 1 - P(no spades)
There are 39 non-spade cards, and so (39 choose 3) = 9139 possible hands not containing spades.
So we find
P(3 spades) = 286/22,100
P(at least 1 spade) = 1 - 9139/22,100 = 12,961/22,100
which means
P(3 spades | at least 1 spade) = 286/12,961 = 0.0221