Đáp án:
$\begin{array}{l}
0 < \alpha < \frac{\pi }{2}\\
a)\pi < \alpha + \pi < \frac{{3\pi }}{2}\\
\Rightarrow \cos \left( {\alpha + \pi } \right) < 0\\
\Rightarrow A < 0\\
b)\\
- \pi < \alpha - \pi < - \frac{\pi }{2}\\
\Rightarrow \left\{ \begin{array}{l}
\sin \left( {\alpha - \pi } \right) < 0\\
\cos \left( {\alpha - \pi } \right) < 0
\end{array} \right.\\
\Rightarrow \tan \left( {\alpha - \pi } \right) > 0\\
\Rightarrow B > 0\\
c)\frac{{2\pi }}{5} < \alpha + \frac{{2\pi }}{5} < \frac{{9\pi }}{{10}}\\
\Rightarrow \sin \left( {\alpha + \frac{{2\pi }}{5}} \right) > 0\\
\Rightarrow C > 0\\
d) - \frac{{3\pi }}{8} < \alpha - \frac{{3\pi }}{8} < \frac{\pi }{8}\\
\Rightarrow \left\{ \begin{array}{l}
\sin \left( {\alpha - \frac{{3\pi }}{8}} \right) < 0\,khi: - \frac{{3\pi }}{8} < x < 0\\
\sin \left( {\alpha - \frac{{3\pi }}{8}} \right) \ge 0\,khi:0 \le x \le \frac{\pi }{8}
\end{array} \right.
\end{array}$
