(1) If the set $M=\left\{x \left\lvert\, \log _{\frac{1}{2}}(x-1)>-1\right.\right\}, N=\left\{x \mid 1<2^{x}<\right.$ $4\}$, then $M \cap N=(\quad)$.
(A) $\{x \mid 1<x<3\}$
(B) $\{x \mid 1<x<2\}$
(C) $\{x \mid 0<x<3\}$
(D) $\{x \mid 0<x<2\}$
(1) B Hint: $x \in M \Leftrightarrow \log _{\frac{1}{2}}(x-1)>-1 \Leftrightarrow\left\{\begin{array}{l}x-1>0, \\ x-1<\left(\frac{1}{2}\right)^{-1}=2\end{array} \Leftrightarrow\right.$ $1<x<3 ; x \in N \Leftrightarrow 1<2^{x}<4 \Leftrightarrow 0<x<2$. Therefore, $M \cap N=\{x \mid$ $1<x<2\}$. Hence, the answer is B.