If A is polynomial time reducible to B and B is in NP, then A is in NP If $A\leq_p B$ and $B$ is in $NP$, is it true that $A$ is in $NP$? 

What about : "if $A\leq_p B$ and $B$ is in $coNP$, then $A$ is in $coNP$"?

Thanks in advance.

I think both hold. If $B$ is in $NP$, then the yes-instances of $B$ can be verified in polynomial time. If $A \leq_p B$, I can "translate" an instance of $A$ to a instance of $B$. Then I can use the verification algorithm for $B$ to check the yes-instances of $A$. The same should be valid for the second affirmation. I'm not sure though.
I think both hold too.

Co-NP version:some Problem A reduce to B, 

if B is co-NP,$\overline{B}$ is in NP,and no-instance of B has some verify way and certificate bit. and A's no-instance can use it too.

So this A is complement of NP problem $\overline{A}$. this means complexity class of A is co-NP.