1. Let the sets be
$$
\begin{array}{l}
S=\left\{x \mid x^{2}-5 x-6<0\right\}, \\
T=\{x|| x+2 \mid \leqslant 3\} .
\end{array}
$$

Then $S \cap T=(\quad)$.
(A) $\{x \mid-5 \leqslant x<-1\}$
(B) $\{x \mid-5 \leqslant x<5\}$
(C) $\{x \mid-1<x \leqslant 1\}$
(D) $\{x \mid 1 \leqslant x<5\}$
-1. C.
Notice,
$$
\begin{array}{l}
S=\{x \mid-1<x<6\}, \\
T=\{x \mid-5 \leqslant x \leqslant 1\} .
\end{array}
$$

Therefore, $S \cap T=\{x \mid-1<x \leqslant 1\}$.