Đáp án:
$\begin{array}{l}
1)\\
\dfrac{\pi }{2} < a < \pi \\
\Rightarrow \sin a > 0\\
\Rightarrow \sin a = \sqrt {1 - {{\cos }^2}a} = \dfrac{4}{5}\\
+ \cos 2a = 2co{s^2}a - 1 = 2.\dfrac{9}{{25}} - 1 = \dfrac{{ - 7}}{{25}}\\
+ \sin 2a = 2.\sin a.\cos a = \dfrac{{ - 12}}{{25}}\\
2)\\
\dfrac{\pi }{2} < x < \pi \\
\Rightarrow \cos x < 0\\
\Rightarrow \cos x = - \sqrt {1 - {{\sin }^2}x} = - \dfrac{3}{5}\\
\Rightarrow \sin 2x = 2.\sin x.\cos x = - \dfrac{{12}}{{25}}\\
\Rightarrow \cos \left( {x - \dfrac{\pi }{3}} \right)\\
= \cos x.cos\dfrac{\pi }{3} + \sin x.\sin \dfrac{\pi }{3}\\
= - \dfrac{3}{5}.\dfrac{1}{2} + \dfrac{4}{5}.\dfrac{{\sqrt 3 }}{2}\\
= \dfrac{{4\sqrt 3 - 3}}{{10}}\\
3)\\
\pi < x < \dfrac{{3\pi }}{2}\\
\Rightarrow \cos x < 0\\
\cos 2x = 2{\cos ^2}x - 1\\
\Rightarrow \cos x = - \sqrt {\dfrac{{\cos 2x + 1}}{2}} = - \dfrac{{2\sqrt 5 }}{5}
\end{array}$
