1. Given $A=\left\{x \left\lvert\, \frac{1}{81}<3^{x-1} \leqslant 3\right., x \in \mathbf{Z}\right\}, B=\left\{x \left\lvert\, \frac{x+2}{x-3}<0\right., x \in \mathbf{N}\right\}$, then the number of elements in the set $C=\{m \mid m=x y, x \in A, y \in B\}$ is $\qquad$
$$
3^{-4}<3^{x-1} \leqslant 3 \Rightarrow -3<x \leqslant 2 \Rightarrow A=\{-2,-1,0,1,2\} ;
$$

Also, $B=\{0,1,2\} \Rightarrow C=\{0, \pm 1, \pm 2, \pm 4\}$, so the number of elements in set $C$ is 7.