Is $AC^0/poly \cap NP$ contained in $P$?
Answer to your first question: Seems unlikely.

Theorem: If $NP \cap AC^0/poly \subseteq P$ then $NEXP = EXP$.

Given a circuit $C$ that outputs a bit, define the decompression of C to be the bit string obtained by evaluating $C$ on all possible inputs. That is, the decompression is $C(0^n) C(0^{n-1}1) C(0^{n-2}10) \cdots C(1^n)$. 

Define the Succinct 3SAT problem as: given a circuit $C$ of size $n$, does its decompression encode a satisfiable Boolean formula? Succinct 3SAT is well-known to be $NEXP$ complete.

Now consider the language 

$L = ${$1^n | $the integer $n$ written in binary is a yes-instance of Succinct 3SAT}.

$L$ is clearly in $AC^0/poly$, since you can just hardcode whether $1^n$ is in $L$, for each $n$.

$L$ is also in $NP$: the integer $n$ written in binary has length about $\log n$, so the decompression of this circuit has length no more than $O(n)$. Hence the satisfying assignment has length at most $O(n)$.

But by the same observations, if $L \in P$, then $NEXP=EXP$, because it means that you 
have an $O(n^c)$ time algorithm for deciding every instance of Succinct 3SAT of length $\log n$.

Your second question is wide open (and open-ended).