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
For each of these, we approach directly via definitions of conditional probability.
$$Pr(A\mid B) = \dfrac{Pr(A\cap B)}{Pr(B)}$$
Here, we calculate the probability that the first two cards are kings as well as the probability that the first two are kings and the third is a queen. The first two are kings happens with probability $\dfrac{4\times 3}{52\times 51}$, seen by direct applications of multiplication principle and counting techniques. Similarly, the probability that the first two are kings and the third is a queen will be $\dfrac{4\times 3\times 4}{52\times 51\times 50}$.
Taking the ratio, the result is going to be $\dfrac{4}{50}$. You might have just skipped straight to the result here, noting that regardless what the two kings happened to be, after they are drawn there will always be four queens left out of 50 cards available.
This one is a bit trickier. We begin the same way, calculating the probability that the first two cards are both spades. This will be $\dfrac{13\times 12}{52\times 51}$, no real difficulty there.
Now, the probability that the third card is a queen and the first two cards are spades... as you correctly noted it is possible that one of the spades were also a queen which would affect our chances of pulling a queen on the third draw. To calculate this, let us choose the cards out of order... beginning by choosing the third card and then choosing the others. We could either choose the queen of spades and then two other spades, or we could choose a different queen followed by two spades. We have then a probability of $\dfrac{1\times 12\times 11 + 3\times 13\times 12}{52\times 51\times 50}$ for the probability of having two spades followed by a queen.
Taking the ratio, we have the conditional probability is then:
$$\dfrac{11+3\times 13}{13\times 50}=\dfrac{1}{13}$$
Interestingly, this is exactly the same as the probability that the third card was a queen., thus proving that the events are in fact independent. It was worth being cautious about this however and not assuming that they were independent before proving it to be the case.
For the third, I would approach similarly to how I did for the second, breaking into cases based on if the spade happened to be an ace or king.