Đáp án:

$\begin{array}{l}
1)Dkxd:x\# 3;y\# \dfrac{3}{2};y\#  - \dfrac{3}{2}\\
\left\{ \begin{array}{l}
\dfrac{8}{{x - 3}} + \dfrac{1}{{2\left| y \right| - 3}} = 5\\
\dfrac{4}{{x - 3}} + \dfrac{1}{{2\left| y \right| - 3}} = 3
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{4}{{x - 3}} = 2\\
\dfrac{1}{{2\left| y \right| - 3}} = 3 - \dfrac{4}{{x - 3}}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x - 3 = 2\\
\dfrac{1}{{2\left| y \right| - 3}} = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 5\left( {tmdk} \right)\\
2\left| y \right| - 3 = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 5\\
\left| y \right| = 2
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 5\\
y = 2/y =  - 2
\end{array} \right.\left( {tmdk} \right)\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {5;2} \right)\left( {5; - 2} \right)} \right\}
\end{array}$

2) a) Xét pt hoành độ giao điểm ta có:

$\begin{array}{l}
{x^2} = mx - m + 2\\
 \Leftrightarrow {x^2} - mx + m - 2 = 0\\
Khi:m = 2\\
 \Leftrightarrow {x^2} - 2x = 0\\
 \Leftrightarrow x.\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0 \Leftrightarrow y = {x^2} = 0\\
x = 2 \Leftrightarrow y = {x^2} = 4
\end{array} \right.\\
 \Leftrightarrow \left( d \right) \cap \left( P \right) = \left( {0;0} \right);\left( {2;4} \right)
\end{array}$

b) 

$\begin{array}{l}
{x^2} - mx + m - 2 = 0\\
\Delta  = {m^2} - 4\left( {m - 2} \right)\\
 = {m^2} - 4m + 8\\
 = {\left( {m - 2} \right)^2} + 4 > 0
\end{array}$

Vậy (d) và (P) luôn cắt nhau tại 2 điểm phân biệt

$\begin{array}{l}
c)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m\\
{x_1}{x_2} = m - 2
\end{array} \right.\\
{x_1} = 2{x_2}\\
 \Leftrightarrow \left\{ \begin{array}{l}
3{x_2} = m\\
{x_1} = 2{x_2}\\
{x_1}{x_2} = m - 2
\end{array} \right.\\
 \Leftrightarrow 2x_2^2 = m - 2\\
 \Leftrightarrow 2.{\left( {\dfrac{m}{3}} \right)^2} = m - 2\\
 \Leftrightarrow \dfrac{{2{m^2}}}{9} = m - 2\\
 \Leftrightarrow 2{m^2} - 9m + 18 = 0\\
 \Leftrightarrow {m^2} - \dfrac{9}{2}m + 9 = 0\left( {vn} \right)
\end{array}$

Vậy ko có m thỏa mãn yêu cầu