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Đáp án:$\begin{array}{l}
a)\sin x = \dfrac{{2\sqrt 5 }}{5};\cos x = \dfrac{{ - \sqrt 5 }}{5};\cot x =  - \dfrac{1}{2}\\
b)A = \dfrac{{385}}{{27}}
\end{array}$

Giải thích các bước giải:

$\begin{array}{l}
a){0^0} < x < {180^0}\\
 \Leftrightarrow \sin x > 0\\
\tan x =  - 2\\
 \Leftrightarrow \cot x = \dfrac{1}{{\tan x}} =  - \dfrac{1}{2}\\
Do:\dfrac{1}{{{{\sin }^2}x}} = {\cot ^2}x + 1 = {\left( { - \dfrac{1}{2}} \right)^2} + 1 = \dfrac{5}{4}\\
 \Leftrightarrow {\sin ^2}x = \dfrac{4}{5}\\
 \Leftrightarrow \sin x = \dfrac{{2\sqrt 5 }}{5}\\
 + \cot x = \dfrac{{\cos x}}{{\sin x}}\\
 \Leftrightarrow \cos x = \cot x.\sin x = \dfrac{{ - 1}}{2}.\dfrac{{2\sqrt 5 }}{5} = \dfrac{{ - \sqrt 5 }}{5}\\
Vay\,\sin x = \dfrac{{2\sqrt 5 }}{5};\cos x = \dfrac{{ - \sqrt 5 }}{5};\cot x =  - \dfrac{1}{2}\\
b)\cot x = \dfrac{3}{5}\\
 \Leftrightarrow \dfrac{{\cos x}}{{\sin x}} = \dfrac{3}{5}\\
 \Leftrightarrow \sin x = \dfrac{5}{3}.\cos x\\
 + \tan x = \dfrac{1}{{\cot x}} = \dfrac{5}{3}\\
 \Leftrightarrow {\tan ^2}x = \dfrac{{25}}{9}\\
 + \dfrac{1}{{{{\cos }^2}x}} = {\tan ^2}x + 1 = \dfrac{{25}}{9} + 1 = \dfrac{{34}}{9}\\
 \Leftrightarrow {\cos ^2}x = \dfrac{9}{{34}}\\
A = \dfrac{{\sin x.\left( {2 + {{\cos }^2}x} \right)}}{{{{\cos }^3}x}}\\
 = \dfrac{{\sin x}}{{\cos x}}.\dfrac{{2 + {{\cos }^2}x}}{{{{\cos }^2}x}}\\
 = \dfrac{5}{3}.\dfrac{{2 + \dfrac{9}{{34}}}}{{\dfrac{9}{{34}}}}\\
 = \dfrac{5}{3}.\dfrac{{77}}{9}\\
 = \dfrac{{385}}{{27}}
\end{array}$