I give an answer in case the modules are finite dimensional (not 100% sure whether the algebra also has to be finite dimensional but I think it is not needed). The base field can be arbitrary and does not need to be $\mathbb{C}$. Let the algebra $A$ be over a field $K$ and $D=Hom_K(-,K)$. Then we have $N \otimes_A M \cong D Hom_A(N,D(M))$.

This makes it very easy to at least calculate the vector space dimension of $N \otimes_A M$ and give alot of examples for Q1: Just take $M$ and $N$ simple such that $N$ is not isomorphic to $D(M)$ and by Schur's lemma $N \otimes_A M \cong D Hom_A(N,D(M))=0$.

Example: Let $A$ be a quiver algebra with $n$ points and simple right modules $S_1,...,S_n$ and simple left modules $G_1,...,G_n$ (corresponding to the points). Then $D(G_i) \cong S_i$ and thus $S_j \otimes_A G_i \neq 0$ if and only if $i=j$.

At least when the algeba is a finite dimensional quiver algebra deciding when $D Hom_A(N,D(M))=0$ can be reduced to linear algebra and answers Q2.

I give an answer in case the modules are finite dimensional (not 100% sure whether the algebra also has to be finite dimensional but I think it is not needed). The base field can be arbitrary and does not need to be $\mathbb{C}$. Let the algebra $A$ be over a field $K$ and $D=Hom_K(-,K)$. Then we have $N \otimes_A M \cong D Hom_A(N,D(M))$.

This makes it very easy to at least calculate the vector space dimension of $N \otimes_A M$ and give alot of examples for Q1: Just take $M$ and $N$ simple such that $N$ is not isomorphic to $D(M)$ and by Schur's lemma $N \otimes_A M \cong D Hom_A(N,D(M))=0$.

Example: Let $A$ be a finite dimensional algebra with $n$ fixed idempotent $e_1,...,e_n$ and simple right modules $S_1,...,S_n$ and simple left modules $G_1,...,G_n$ (corresponding to the idempotents ). Then $D(G_i) \cong S_i$ and thus $S_j \otimes_A G_i \neq 0$ if and only if $i=j$.

At least when the algeba is a finite dimensional quiver algebra deciding when $D Hom_A(N,D(M))=0$ can be reduced to linear algebra and answers Q2.