Assuming a scenario where opponent holds exactly AA, and I hold AK.

What is the percentage that a flop will contain the last ace?
...that a flop will contain one of the 3 kings?

I'll give it a shot. Please confirm or correct:

Probability for flopping an ace:
C(50;3) - C(49;3) = 1176
1176/19600 = 0,06
= 6%

Probability for flopping one of 3 kings:
C(50;3) - C(47;3) = 3385
3385/19600 = 0,1727
= 17,27%

Combined:
= 6% + 17,27%
= 23,27%

You know your opponent's cards and yours, so that leaves only 48 cards to draw from for the flop.
C(48,3) - C(47,3) = 17296 - 16215 = 1081
1081/17296 = .0625 or 6.25%

C(48,3) - C(45,3) = 17296 - 14190 = 3106
3106/17296 = .1796 or 17.96%

Adding the probabilities isn't valid because they're not mutually exclusive (some flops contain both the ace and king(s)). Instead, just do the same as the earlier calculations except now any of 4 cards on the flop qualifies (1 ace + 3 kings):

C(48,3) - C(44,3) = 17296 - 13244 = 4052
4052/17296 = .2343 or 23.43%

The reason your answer ended up being so close is because two errors partially cancelled each other out. The base probabilities were slightly low, and summing probabilities on events that aren't mutually exclusive always overshoots the true probability.

Thanks

Note that the probability of flopping an A is obviously 3/48 since each of the flop cards has a 1/48 chance, and since there is only 1 ace remaining they are mutually exclusive, so we can just multiply by 3.