We have by the definition of conditional probability
$$\Pr(E_2|E_1)=\frac{\Pr(E_1\cap E_2)}{\Pr(E_1)}.$$
We calculate the two probabilities on the right-hand side.
The probability of $E_1$ is $1$ minus the probability of no Kings. You calculated $\Pr(E_1)$ correctly.
For the probability of $E_1\cap E_2$, which is just the probability of $E_2$, use the same basic strategy. The number of $1$ King hands is $\binom{4}{1}\binom{48}{4}$, for we have to choose $4$ non-Kings to go with the King. So the probability of at least $2$ Kings is
$$1-\frac{\binom{4}{0}\binom{48}{5}+\binom{4}{1}\binom{48}{4}}{\binom{52}{5}}.$$
Without Replacement: You shuffle the deck thoroughly, take out three cards. For this particular problem, the question is "What is the probability these cards are all Kings."
With Replacement: Shuffle the deck, pick out one card, record what you got. Then put it back in the deck, shuffle, pick out one card, record what you got. Then put it back in the deck, pick out one card, record what you got. One might then ask for the probability that all three recorded cards were Kings. In the with replacement situation, it is possible, for example, to get the $\spadesuit$ King, or the $\diamondsuit$ Jack more than once.
For solving the "without replacement" problem, here are a couple of ways. There are $\binom{52}{3}$ equally likely ways to choose $3$ cards. There are $\binom{4}{3}$ ways to choose $3$ Kings. So our probability is $\binom{4}{3}/\binom{52}{3}$.
Or else imagine taking out the cards one at a time. The probability the first card taken out was a King is $\frac{4}{52}$. Given that the first card taken out was a King, the probability the second one was is $\frac{3}{51}$, since there are $51$ cards left of which $3$ are Kings. So the probability the first two cards were Kings is $\frac{4}{52}\cdot\frac{3}{51}$. **Given that the first two were Kings, the probability the third is is $\frac{2}{50}$. So the desired probability is $\frac{4}{52}\cdot\frac{3}{51}\cdot \frac{2}{50}$.
Remark: We could solve the same three Kings problem under the "with replacement" condition. (You were not asked to do that,) The second approach we took above yields the answer $\left(\frac{4}{52}\right)^3$. Since we are replacing the card each time and shuffling, the probability of what the "next" card is is not changed by the knowledge that the first card was a King.
Best Answer
There are $\binom{52}{13}$ ways to select 13 cards from 52 cards.
Consider how you would create a hand of 13 cards with the required number of kings.
There are $\binom{4}{n}$ ways of choosing the required number of kings.
There are then $\binom{48}{13-n}$ ways of choosing the remaining (13 - n) cards.
Let A be the event of getting at least 2 kings.
This can be thought of as 3 separate events corresponding to 2, 3 or 4 kings. By the addition principle:
Number of ways of getting A = $\binom{4}{2}$ * $\binom{48}{13-2}$ + $\binom{4}{3}$ * $\binom{48}{13-3}$ + $\binom{4}{4}$ * $\binom{48}{13-4}$
So, $$P(A) = \frac{\binom{4}{2} * \binom{48}{13-2} + \binom{4}{3} * \binom{48}{13-3} + \binom{4}{4} * \binom{48}{13-4}}{\binom{52}{13}}$$
Let B be the event of getting at least 3 kings.
This can be thought of as 2 separate events corresponding to 3 or 4 kings. By the addition principle:
$$P(B) = \frac{\binom{4}{3} * \binom{48}{13-3} + \binom{4}{4} * \binom{48}{13-4}}{\binom{52}{13}}$$
Finally, by the conditional probability formula:
$$P(B|A) = \frac{P(B \cap A)}{P(A)}= \frac{P(B)}{P(A)}$$