Your first answer is correct, but overkill.
You can rephrase the question as "What is the probability that the first honor card (ace,king,queen, or jack) is an ace." Then it is clearly 1/4 - each is equally likely.
Note that all the probabilities are the same if you remove the cards not in question - so that you can think of it as a deck of 16 cards. There are $16!$ orders for the deck. There are $3!4^4$ different ways to have the first four honors be all different and the fourth be an ace.
So the probability for the second case is:
$$\frac{3!4^412!}{16!}\approx 0.035$$
Which is significantly higher than your probability.
Did you assume that there were no cards other than the king, queen, and jack before the ace? Then you need all 52 cards, and you'd get:
$$\frac{3!4^448!}{52!}=\frac{3!4^4}{52\cdot 51\cdot 50\cdot 49}\approx 0.000236$$
Which is closer to your result, but still different.
For all three of these questions, the proper answer depends on applying Bayes' Theorem correctly:
$$
P(A|B) * P(B) = P(B|A) * P(A)
$$
Where $A$ is the event "draw two queens and a king", and $B$ is the specific known condition in each part. First note that in all three cases, $P(B|A)$ is $1$ since drawing two queens and a king satisfies each given condition. Since the problem asks for $P(A|B)$, we'll use:
$$
P(A|B) = \frac{P(A)}{P(B)}
$$
So then, what's $P(A)$? (Since that's the same in all three cases)
Well, drawing two queens and then a king does indeed have a probability of
$$
\left(\frac{4}{52}\right)^3
$$
But "two queens and a king" would imply that the king can be drawn first, second, or third so in fact
$$
P(A) = 3 \left(\frac{4}{52}\right)^3
$$
So now for the the parts:
In part (a) $B$ is "at least one queen". The chance of "at least one queen" is going to be $1$ minus the chance of "no queens", so:
$$
P(A|B) = \frac{3 \left(\frac{4}{52}\right)^3}{1 - \left(\frac{48}{52}\right)^3} = \frac{3}{469}
$$
In part (b), $B$ is "at least two face cards". There a few ways to calculate $P(B)$, but I think that the easiest conceptually is to split it into two (mutually exclusive) cases: two face cards and a non-face card ($B_1$), or three face cards.($B_2$)
In the first case, the non-face card can be first, second, or third, so
$$
P(B_1) = 3 \left(\frac{12}{52}\right)^2\left(\frac{40}{52}\right) = \frac{270}{2197}
$$
And in the second case:
$$
P(B_2)= \left(\frac{12}{52}\right)^3 = \frac{27}{2197}
$$
So:
$$
P(A|B) = \frac{3 \left(\frac{4}{52}\right)^3}{\frac{27}{2197} + \frac{270}{2197}} = \frac{1}{99}
$$
In part (c), $B$ is "all cards are greater than 7", so that's saying that every card is 8, 9, 10, J, Q, or K. (I'm assuming here that Ace is counted as lowest card, not highest) so $P(B)$ is just $(24/52)^3$, and
$$
P(A|B) = \frac{3 \left(\frac{4}{52}\right)^3}{\left(\frac{24}{52}\right)^3} = \frac{1}{72}
$$
Best Answer
Our mini-deck has $12$ cards. There are $\binom{12}{4}$ equally likely ways to choose $4$ cards from this deck.
Now we count the favourables, the hands that have $2$ Queens and $2$ Kings.
The $2$ Queens can be chosen from the $4$ available in $\binom{4}{2}$ ways. For each way of choosing $2$ Queens, there are $\binom{4}{2}$ ways to choose the $2$ Kings. Thus the total number of favourables is $\binom{4}{2}\binom{4}{2}$.
For the probability, divide.
For the second problem, there are $\binom{4}{2}$ ways to choose the $2$ Queens. Once this is done, the Kings are determined, since they must be of the same suits as the chosen Queens. Thus the number of favourables in the "suits must match" problem is just $\binom{4}{2}$.