What is the probability of getting more than 4 heads with 6 coins given that you got at least 2 heads?

If you roll five dice what is the probability of getting exactly three dice with a "4" and two something with other than a "4"?

"What is the probability of getting more than 4 heads with 6 coins given that you got at least 2 heads?"

This is conditional probability: put
A:= Event "getting more than 4 heads"
B:= Event "got at least 2 heads"

then what are you looking is P(A|B)=\frac{P(A\cap B)}{P(B)}.
Now it is clear that A\subset B (as if A is verified, then B is verified too), so what you are looking for becomes \frac{P(A)}{P(B)}.

Event "getting more than 4 heads": you can want either 5 heads (prob. 6/64) or 6 heads (prob. 1/64).
Event "got at least 2 heads": it is the complementary of event "getting more than 4 crosses".

"If you roll five dice what is the probability of getting exactly three dice with a "4" and two something with other than a "4"?"

You have to choose 3 dices in which you get "4": you can do this in \frac{5!}{3!2!} ways, and the probability of getting "4" is 1/6. Then the other 2 dices must not be "4", which has probability 5/6. Then multiply them.