1. Let the set $A=\left\{x \left\lvert\, \frac{x+4}{x-3} \leqslant 0\right., x \in \mathbf{Z}\right\}$, and from the set $A$ a random element $x$ is drawn, denoted by $\xi=x^{2}$. Then the mathematical expectation of the random variable $\xi$ is $\mathrm{E} \xi=$ $\qquad$
$-, 1.5$
From the conditions, we know
$$
A=\{-4,-3,-2,-1,0,1,2\},
$$

The values of the random variable $\xi$ are $0,1,4,9,16$.
It is easy to see that the probability distribution of $\xi$ is shown in Table 1. Therefore, E $\xi$
$$
\begin{array}{l}
=0 \times \frac{1}{7}+1 \times \frac{2}{7}+4 \times \frac{2}{7}+9 \times \frac{1}{7}+16 \times \frac{1}{7} \\
=5
\end{array}
$$