If A is not in NP, and A reduces to B, does this mean B is not in NP?
I'm assuming that both $A$ and $B$ are decision problems and that we are talking bout Karp reductions.
Suppose towards a contradiction that $A \not\in NP$, $A \le_p B$, and $B \in NP$.
Then, a non-deterministic polynomial-time Turing machine that decides $A$ would be the following:

Use the Karp reduction $f$ from $A$ to $B$ to transform an instance $x$ of $A$ into an instance $f(x)$ of $B$;
Simulate a non-deterministic polynomial-time Turing machine $T$ that decides $B$ on input $f(x)$ ($T$ exists since $B \in NP$);
If $T(x)$ accepts, accept. If $T(x)$ rejects, reject.

This is a contradiction.