Let $M, N$ be $R$-modules and $(\cdot, \cdot): M \times N \to R$ be a perfect pairing. Wikipedia sais that this means that the map $\varphi: M \to \text{Hom}_R(N, R), m \mapsto (n \mapsto (m, n))$ is an isomorphism.
Question 1: This question is as important as the others, so it would be sufficient for me if you say if it's true or not: Does this definition of "perfect" imply that the other mapping $N \to \text{Hom}_R(M, R), n \mapsto (m \mapsto (m, n))$ is an isomorphism?
Now, let $P$ be another $R$-module.
Question 2: Does the perfect pairing in general induce an isomorphism $N \otimes_R P \to \text{Hom}_R(M, P)$?
I wanted to define a possible isomorphism via $n \otimes p \mapsto (m \mapsto (m, n)\cdot p)$ but I have trouble finding an inverse. If this is not true, then perhaps the following weaker statement?
Question 3: Assume additionally that $R = \mathbb{Z}$ and that $M$ and $N$ are lattices and $P = \mathbb{C}^*$. Is statement $2$ then true?
Assuming your definition of perfect pairing, the answer to your first question is no. You are essentially asking if $N \cong\text{Hom}(\text{Hom}(N,R), R))$ and this is not true when $N = \mathbb{Z}/2$, $M = 0$ and $R = \mathbb{Z}$.
The same example with $P = \mathbb{Z}$ tells you that the answer to your second question must also be no.
I don't understand how $\mathbb{C}^*$ is a $\mathbb{Z}$-module, so your third question doesn't make sense to me. Are you using some sort of exponential map?