Đáp án:

$\begin{array}{l}
a)DKxd:x \ne 3;x \ne  - 3\\
A = \left( {\frac{x}{{x + 3}} - \frac{2}{{x - 3}} + \frac{{{x^2} - 1}}{{9 - {x^2}}}} \right):\left( {2 - \frac{{x + 5}}{{3 + x}}} \right)\\
 = \frac{{x\left( {x - 3} \right) - 2\left( {x + 3} \right) - \left( {{x^2} - 1} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}:\left( {\frac{{2\left( {3 + x} \right) - x - 5}}{{x + 3}}} \right)\\
 = \frac{{{x^2} - 3x - 2x - 6 - {x^2} + 1}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\frac{{x + 3}}{{6 + 2x - x - 5}}\\
 = \frac{{ - 5x - 5}}{{x - 3}}.\frac{1}{{x + 1}}\\
 = \frac{{ - 5}}{{x - 3}}\\
 = \frac{5}{{3 - x}}\\
b){x^2} - x - 2 = 0\\
 \Leftrightarrow {x^2} - 2x + x - 2 = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\left( {tm} \right)\\
x =  - 1\left( {tm} \right)
\end{array} \right.\\
 + Khi:x = 2 \Rightarrow A = \frac{5}{{3 - x}} = 5\\
 + Khi:x =  - 1 \Rightarrow A = \frac{5}{{3 - x}} = \frac{5}{4}\\
Vậy\,A = 5/A = \frac{5}{4}\,khi:{x^2} - x - 2 = 0\\
c)A = \frac{1}{2}\\
 \Rightarrow \frac{5}{{3 - x}} = \frac{1}{2}\\
 \Rightarrow 3 - x = 10\\
 \Rightarrow x =  - 7\left( {tm} \right)\\
Vậy\,x =  - 7
\end{array}$