Đáp án:

$\begin{array}{l}
1)a)Dkxd:x \ne 1;x \ne  - 2\\
P = \left( {\frac{{x - 2}}{{x - 1}} - \frac{x}{{x + 2}}} \right):\frac{1}{{x + 2}}\\
 = \frac{{\left( {x - 2} \right)\left( {x + 2} \right) - x\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)}}.\left( {x + 2} \right)\\
 = \frac{{{x^2} - 4 - {x^2} + x}}{{x - 1}}\\
 = \frac{{x - 4}}{{x - 1}}\\
b)x =  - \frac{1}{2}\left( {tmdk} \right)\\
 \Rightarrow P = \frac{{x - 4}}{{x - 1}} = \frac{{\frac{{ - 1}}{2} - 4}}{{ - \frac{1}{2} - 1}} = 3\\
c)P > 1\\
 \Rightarrow \frac{{x - 4}}{{x - 1}} > 1\\
 \Rightarrow \frac{{x - 4 - x + 1}}{{x - 1}} > 0\\
 \Rightarrow \frac{{ - 3}}{{x - 1}} > 0\\
 \Rightarrow x - 1 < 0\\
 \Rightarrow x < 1\\
Vay\,x < 1;x \ne  - 2\,thi\,P > 1\\
2)a){x^2} = 4x\\
 \Rightarrow {x^2} - 4x = 0\\
 \Rightarrow x\left( {x - 4} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 4
\end{array} \right.\\
\frac{{x + 3}}{{x - 1}} - \frac{3}{{x + 2}} = \frac{{14 - 2x}}{{{x^2} + x - 2}}\left( {dk:{x_1};x \ne  - 2} \right)\\
 \Rightarrow \frac{{\left( {x + 3} \right)\left( {x + 2} \right) - 3\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)}} = \frac{{14 - 2x}}{{\left( {x - 1} \right)\left( {x + 2} \right)}}\\
 \Rightarrow {x^2} + 5x + 6 - 3x + 3 = 14 - 2x\\
 \Rightarrow {x^2} + 4x - 5 = 0\\
 \Rightarrow \left( {x - 1} \right)\left( {x + 5} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 1\left( {ktm} \right)\\
x =  - 5\left( {tm} \right)
\end{array} \right.\\
Vay\,x =  - 5\\
b)pt:{x^2} - 3x + m - 5 = 0\,co\,x = 0;x = 4\\
 + Khi:x = 0 \Rightarrow m - 5 = 0 \Rightarrow m = 5\\
 + khi:x = 4 \Rightarrow 16 - 12 + m - 5 = 0 \Rightarrow m = 1
\end{array}$

Vậy m=5 hoặc m=1