Đáp án:

$\begin{array}{l}
1){\mathop{\rm cota}\nolimits}  =  - 2\\
 \Rightarrow tana =  - \dfrac{1}{2}\\
Do:\dfrac{1}{{{{\cos }^2}a}} = {\tan ^2}a + 1 = \dfrac{5}{4}\\
 \Rightarrow {\cos ^2}a = \dfrac{4}{5}\\
 \Rightarrow \cos a =  - \dfrac{{2\sqrt 5 }}{5}\left( {do:\dfrac{\pi }{2} < a < \pi } \right)\\
 \Rightarrow \sin a = \dfrac{{\sqrt 5 }}{5}\\
2)\cos a =  - \dfrac{3}{5}\\
 \Rightarrow \sin a =  - \sqrt {1 - {{\cos }^2}a}  = \dfrac{{ - 4}}{5}\\
 \Rightarrow \tan a = \dfrac{{\sin a}}{{\cos a}} = \dfrac{4}{3}\\
 \Rightarrow \cot a = \dfrac{3}{4}\\
3)\sin a =  - \dfrac{{12}}{3} =  - 4\left( {ktm} \right)\\
4)\tan a = \dfrac{4}{{15}}\\
 \Rightarrow \sin a = \dfrac{4}{{15}}.\cos a\\
 \Rightarrow A = \dfrac{{5\sin a + 7\cos a}}{{6\cos a - 3\sin a}}\\
 = \dfrac{{5.\dfrac{4}{{15}}.{\mathop{\rm cosa}\nolimits}  + 7cosa}}{{6.\cos a - 3.\dfrac{4}{{15}}.\cos a}}\\
 = \dfrac{{125}}{{123}}\\
5){\mathop{\rm tana}\nolimits}  = 2 \Rightarrow sina = 2cosa\\
 \Rightarrow \dfrac{{8{{\cos }^3}a - 2{{\sin }^3}a + \cos a}}{{2\cos a - {{\sin }^2}a}}\\
 = \dfrac{{8{{\cos }^2}a - 16{{\cos }^3}a + \cos a}}{{2.\cos a - 8{{\cos }^3}a}}\\
 = 1 - \dfrac{1}{{2 - 8{{\cos }^2}a}}\\
2a)\\
A = \dfrac{{1 + {\mathop{\rm sinx}\nolimits} }}{{\sin \dfrac{x}{2} + \cos \dfrac{x}{2}}} - 2{\cos ^2}\dfrac{x}{4}\\
 = \dfrac{{{{\sin }^2}\dfrac{x}{2} + 2.\sin \dfrac{x}{2}.\cos \dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2}}}{{\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)}} - 2{\cos ^2}\dfrac{x}{4}\\
 = \dfrac{{{{\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)}^2}}}{{\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)}} - \left( {2{{\cos }^2}\dfrac{x}{4} - 1} \right) - 1\\
 = \left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right) - \cos \dfrac{x}{2} - 1\\
 = \sin \dfrac{x}{2} - 1\\
B = 3 - 4\cos 2a + \cos 4a\\
 = 3 - 4\cos 2a + 2{\cos ^2}2a - 1\\
 = 2{\cos ^2}2a - 4\cos 2a + 2\\
 = 2.{\left( {\cos 2a - 1} \right)^2}\\
 = 2.{\left( {{{\sin }^2}a} \right)^2}\\
 = 2.{\sin ^4}a
\end{array}$