Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
0 < a < \frac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}
\sin a > 0\\
\cos a > 0
\end{array} \right.\\
\cos a > 0 \Rightarrow \cos a = \sqrt {1 - {{\sin }^2}a}  = \frac{{\sqrt {15} }}{4}\\
\sin 2a = 2\sin a.\cos a = \frac{{2\sqrt {15} }}{{16}}\\
\cos 2a = 1 - 2{\sin ^2}a = \frac{7}{8}\\
\tan 2a = \frac{{\sin 2a}}{{\cos 2a}} = \frac{{\sqrt {15} }}{7}\\
2,\\
\cot b = \frac{4}{3} \Rightarrow \tan b = \frac{3}{4}\\
\tan \left( {a + b} \right) = \frac{{\tan a + \tan b}}{{1 - \tan a.\tan b}} = \frac{{\frac{1}{7} + \frac{3}{4}}}{{1 - \frac{1}{7}.\frac{3}{4}}} = 1 \Rightarrow a + b = 45^\circ \\
3,\\
{\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8}\\
 = {\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\left( {\pi  - \frac{{5\pi }}{8}} \right) + {\sin ^4}\left( {\pi  - \frac{{7\pi }}{8}} \right)\\
 = 2\left[ {{{\sin }^4}\frac{\pi }{8} + {{\sin }^4}\frac{{3\pi }}{8}} \right]\\
 = 2\left[ {{{\sin }^4}\frac{\pi }{8} + {{\cos }^4}\left( {\frac{\pi }{2} - \frac{{3\pi }}{8}} \right)} \right]\\
 = 2.\left( {{{\sin }^4}\frac{\pi }{8} + {{\cos }^4}\frac{\pi }{8}} \right)\\
 = 2.\left[ {{{\left( {{{\sin }^2}\frac{\pi }{8} + {{\cos }^2}\frac{\pi }{8}} \right)}^2} - 2.{{\sin }^2}\frac{\pi }{8}.{{\cos }^2}\frac{\pi }{8}} \right]\\
 = 2.\left[ {1 - \frac{1}{2}.{{\sin }^2}\frac{\pi }{4}} \right]\\
 = \frac{3}{2}\\
4,\\
\sin a = \frac{1}{3} \Leftrightarrow 2\sin \frac{a}{2}.\cos \frac{a}{2} = \frac{1}{3}\\
{\left( {\sin \frac{a}{2} + \cos \frac{a}{2}} \right)^2} = {\sin ^2}\frac{a}{2} + 2.\sin \frac{a}{2}.\cos \frac{a}{2} + {\cos ^2}\frac{a}{2} = 1 + \frac{1}{3} = \frac{4}{3}\\
0 < a < \frac{\pi }{2} \Rightarrow \sin \frac{a}{2} + \cos \frac{a}{2} > 0\\
 \Rightarrow \sin \frac{a}{2} + \cos \frac{a}{2} = \frac{2}{{\sqrt 3 }}\\
5,\\
2{\sin ^2}x - 5\sin x + 2 = 0\\
 \Leftrightarrow \left( {2\sin x - 1} \right)\left( {\sin x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = \frac{1}{2}\\
\sin x = 2\,\,\,\,\,\left( L \right)
\end{array} \right. \Rightarrow \sin x = \frac{1}{2}\\
\cos 2x = 1 - 2{\sin ^2}x = \frac{1}{2}
\end{array}\)