(1) Let the set $S=\left\{x \mid x^{2}-5 x-6<0\right\}, T=\{x|| x+2 \mid \leqslant 3\}$, 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) $\mathrm{C}$ Hint: $S=\{x \mid-1<x<6\}, T=\{x \mid-5 \leqslant x \leqslant 1\}$, so $S \cap T=\{x \mid-1<x \leqslant 1\}$.