Đáp án:
$\begin{array}{l}
1)Do:\left\{ \begin{array}{l}
- 1 \le \sin a \le 1\\
- 1 \le {\mathop{\rm cosa}\nolimits} \le 1
\end{array} \right.\\
a) - 1 \le \sin \left( {2x + \dfrac{\pi }{3}} \right) \le 1\\
\Rightarrow - 5 \le - 5\sin \left( {2x + \dfrac{\pi }{3}} \right) \le 5\\
\Rightarrow - 1 \le 4 - 5\sin \left( {2x + \dfrac{\pi }{3}} \right) \le 9\\
\Rightarrow - 1 \le y \le 9\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = - 1\\
Khi:\sin \left( {2x + \dfrac{\pi }{3}} \right) = 1\\
\Rightarrow \left( {2x + \dfrac{\pi }{3}} \right) = \dfrac{\pi }{2} + k\pi \\
\Rightarrow x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}\\
GTLN:y = 9\\
Khi:\sin \left( {2x + \dfrac{\pi }{3}} \right) = - 1\\
\Rightarrow 2x + \dfrac{\pi }{3} = \dfrac{{ - \pi }}{2} + k\pi \\
\Rightarrow x = \dfrac{{ - 5\pi }}{{12}} + \dfrac{{k\pi }}{2}
\end{array} \right.\\
b)y = 3 - \sqrt {5\cos 3x + 1} \\
Dkxd:5\cos 3x + 1 \ge 0\\
\Rightarrow \cos 3x \ge - \dfrac{1}{5}\\
\Rightarrow 0 \le 5\cos 3x + 1 \le 6\\
\Rightarrow 0 \le \sqrt {5\cos 3x + 1} \le \sqrt 6 \\
\Rightarrow 3 - \sqrt 6 \le 3 - \sqrt {5\cos 3x + 1} \le 3\\
\Rightarrow 3 - \sqrt 6 \le y \le 3\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = 3 - \sqrt 6 \\
Khi:\cos 3x = 1\\
\Rightarrow x = \dfrac{{k2\pi }}{3}\\
GTLN:y = 3\\
Khi:\cos 3x = 0\\
\Rightarrow x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}
\end{array} \right.\\
2)\\
a)f\left( x \right) = y = \sqrt {\dfrac{{1 + \sin x}}{{1 - \cos x}}} \\
\Rightarrow f\left( { - x} \right) = \sqrt {\dfrac{{1 - \sin x}}{{1 - \cos x}}} \ne - f\left( x \right)\\
\Rightarrow hs\,ko\,chan,ko\,le\\
b)y = {\cot ^2}x + \cos x\\
\Rightarrow f\left( { - x} \right) = {\cot ^2}x + \cos x = f\left( x \right)\\
\Rightarrow hs\,chan
\end{array}$
