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.
There are indeed $\binom{52}{13}$ different 13-card hands and this will indeed be the size of our sample space and thus our denominator when we finish our calculations.
For the numerator, we need to pause for a moment and understand what the problem is actually asking, since this appears to be where you got stuck.
We are asked to find the probability that in our hand of thirteen cards, there is at least one suit for which we have all three face cards. For example $(A\spadesuit,2\spadesuit,3\spadesuit,\dots,10\spadesuit,J\spadesuit,Q\spadesuit,K\spadesuit)$ has all three of the face cards for spades. Similarly if all those cards happened to be hearts instead it would also count since we would have all of the face cards for hearts. Similarly still, a hand like $(J\spadesuit,Q\spadesuit,K\spadesuit,J\heartsuit,Q\heartsuit,K\heartsuit,J\diamondsuit,Q\diamondsuit,K\diamondsuit,\dots)$ would count since we have all of the face cards from spades (we also happen to have all of the face cards from hearts and diamonds too).
Let $\spadesuit$ represent the event that we have have all of the face cards from spades. Similarly, let $\diamondsuit, \heartsuit, \clubsuit$ represent the event that we have all of the face cards from diamonds, hearts, and clubs respectively.
You are asked to find $Pr(\spadesuit\cup\diamondsuit\cup \clubsuit\cup \heartsuit)$
To do this, let us apply inclusion exclusion. We expand the above as:
$Pr(\spadesuit\cup \diamondsuit\cup\clubsuit\cup\heartsuit) = Pr(\spadesuit)+Pr(\diamondsuit)+\dots-Pr(\spadesuit\cap \diamondsuit)-Pr(\spadesuit\cap \clubsuit)-\dots+Pr(\spadesuit\cap \diamondsuit\cap \clubsuit)+\dots-Pr(\spadesuit\cap\diamondsuit\cap \clubsuit\cap \heartsuit)$
Now, let us calculate each individual term in the expansion.
The calculation you did before is relevant. Indeed, we calculate $Pr(\spadesuit)=\dfrac{\binom{3}{3}\binom{49}{10}}{\binom{52}{13}}$. This is again merely the probability that we have all of the face cards from the spades and is not the final probability that we were tasked with calculating.
We continue and calculate more terms:
For example $Pr(\spadesuit\cap \diamondsuit)=\dfrac{\binom{6}{6}\binom{46}{7}}{\binom{52}{13}}$
We then notice what symmetry there is in the terms and can simplify some. Finally, we write the final expression for our final answer (and get an exact number only if actually requested or required, usually opting to leave the answer in terms of binomial coefficients without additional simplification).
Best Answer
We can multiply together the probabilities that have each successive draw satisfy the condition. $$\frac{52}{52}\cdot \frac{48}{51}\cdot\frac{44}{50}\cdot\frac{40}{49}\cdot\frac{36}{48}=\frac{2112}{4165}\approx0.50708$$ On the first draw, you have a probability of $\frac{52}{52}$ of selecting a card that is different from a card in your hand (naturally, as you have no cards in your hand).
On the second draw, you have a probability of $\frac{48}{51}$ of selecting a card that is different from the card you had just drawn (that is, only $\frac3{51}$ match the card in your hand).
Apply for the next three draws.
To get at least one Ace, we must fail the action of getting no Aces. $$1-\frac{48}{52}\cdot\frac{47}{51}\cdot\frac{46}{50}\cdot\frac{45}{49}\cdot\frac{44}{48}=1-\frac{35673}{54145}\approx0.34116$$ Alternatively, we can think of this as the number of ways to choose $5$ from $48$ (non-Aces), out of the number of ways to choose $5$ from $52$ (total cards in deck): $$1-\frac{\binom{48}5}{\binom{52}5}=1-\frac{\frac{48!}{5!\ (48-5)!}}{\frac{52!}{5!\ (52-5)!}}=1-\frac{48!\cdot47!}{43!\cdot52!}=1-\frac{48}{52}\cdot\frac{47}{51}\cdot\frac{46}{50}\cdot\frac{45}{49}\cdot\frac{44}{48}=1-\frac{35673}{54145}\approx0.34116$$ The same calculation! Seems the suits did not have anything to do with the probabilities, after all!