The probability that you will have at most 3 kings is the probability that you will have less than 4.
$$\mathsf P(K\leq 3) = 1 -\mathsf P(K=4)$$
The probability that you will have exactly all four kings is the count of ways to select 4 kings and 1 other card divided by the count of ways to select any 5 cards.
$$\mathsf P(K=4)~=~\dfrac{{^{4}\mathrm C_{4}}\cdot{^{(52-4)}\mathrm C_{(5-4)}}}{^{52}\mathrm C_5}$$
Put it together.
$$\mathsf P(K\leq 3) ~=~ 1 -\dfrac{{^{4}\mathrm C_{4}}\cdot{^{(52-4)}\mathrm C_{(5-4)}}}{^{52}\mathrm C_5}$$
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
There are several approaches to this problem. One of the simplest to explain is that you can make a distribution chart to describe the scenario:
$$\begin{array}{c|c|c|c|c|c|c|} & \color{blue}{1} & \color{red}{1} & \color{green}{1} & \color{yellow}{1}&\color{blue}{2} & \color{red}{2}&\cdots\\ \hline \color{blue}{1} & 0 & x & x & \cdots\\ \hline \color{red}{1} & x & 0 & x & \cdots\\ \hline \color{green}{1} & x & x & 0 & x & \cdots\\ \hline \color{yellow}{1}&\vdots & \ddots & \ddots & \ddots\\ \hline \color{blue}{2} \\ \hline \color{red}{2}\\ \hline \vdots\\ \end{array}$$
(note: these probabilities have it such that order matters. The final answer in this problem will not depend on whether you use a method where order matters or doesn't matter, so use what is comfortable)
Noting that it is impossible to draw the same card twice in a row (hence the main diagonal being zeroes) and all other entries are equiprobable and should add up to 1, you calculate $Pr(\text{first card is}~ a\cap \text{second card is}~ b) = \frac{1}{16\cdot 15}$. You may then note that there are $48$ squares with nonzero probability that correspond to having the same number in both the first draw and the second draw. Use addition principle to complete the argument for part (a). For part (b), add up the probability of all squares whose corresponding outcomes have the total sum is bigger than 6.
That approach is rather tedious and would require either drawing a very large chart, or only drawing a portion of it (like I did) and "reading" parts of the chart that you haven't written down yet. Instead, let us consider this via multiplication and addition principles.
$Pr(\text{first two numbers match}) = Pr(\text{both numbers are 1}\cup \text{both numbers are 2}\cup\text{both numbers are 3}\cup\text{both numbers are 4})\\ =Pr(\text{both numbers are 1}) + Pr(\text{both numbers are 2}) + Pr(\text{both numbers are 3}) + Pr(\text{both numbers are 4})$
Here, I was able to split up the unions ($\cup$) by the addition principle:
Now, we wish to solve for $Pr(\text{both numbers are 1})$. This is the same as $Pr(\text{first is a 1}\cap \text{second is a 1})$. To do this, we use the multiplication rule:
So, we look at what $Pr(\text{first is a 1})$ is and what $Pr(\text{second is a 1}|\text{first is a 1})$. The probability the first card is a 1 is simply $\frac{4}{16}$ (since there are four 1's out of sixteen total cards) and the probability of the second being a 1 given that the first is a one as well is $\frac{3}{15}$ (since there will be three remaining 1's out of fifteen remaining cards total).
Thus, we see $Pr(\text{both are 1}) = \frac{4}{16}\cdot\frac{3}{15}$. Through a similar argument we find that the rest of the probabilities are also $\frac{4}{16}\cdot\frac{3}{15}$ as well. So, $Pr(\text{both numbers the same}) = 4\cdot Pr(\text{both are 1}) = 4\cdot \frac{4}{16}\cdot \frac{3}{15} = \frac{3}{15}$
Once you become more comfortable with calculating probabilities, you may make simplifications which save a great deal more time. As mentioned as a hint in the comments, it doesn't actually matter what the first number drawn is. As such $Pr(\text{both are same number}) = Pr(\text{second number is same as first}) = \frac{3}{15}$ since whatever number the first card was, there will be three remaining of that number out of 15 cards to take from.
For part (b), use the tools described here and note that $Pr(\text{sum of cards is more than 6}) = Pr(\text{sum of cards is 7}\cup \text{sum of cards is 8})$. Note further that you can break $Pr(\text{sum of cards is 7})$ up as $Pr(\text{first card is 3 and second is 4}\cup\text{first card is 4 and second is 3})$.