1. Let the set $A=\left\{x \left\lvert\, \frac{x+4}{x-3} \leqslant 0\right., x \in \mathbf{Z}\right\}$, and from 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 that
$$
A=\{-4,-3,-2,-1,0,1,2\},
$$

The values of the random variable $\xi$ are $0, 1, 4, 9, 16$.
It is easy to obtain that the probability distribution of $\xi$ is shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|}
\hline$\xi$ & 0 & 1 & 4 & 9 & 16 \\
\hline$P$ & $\frac{1}{7}$ & $\frac{2}{7}$ & $\frac{2}{7}$ & $\frac{1}{7}$ & $\frac{1}{7}$ \\
\hline
\end{tabular}

Therefore, $\mathrm{E} \xi$
$$
\begin{array}{l}
=0 \times \frac{1}{7}+1 \times \frac{2}{7}+4 \times \frac{2}{7}+9 \times \frac{1}{7}+16 \times \frac{1}{7} \\
=5 .
\end{array}
$$