Đáp án:

$\begin{array}{l}
1)pt\,BC:y = a.x + b\\
Do:B\left( {0;3} \right);C\left( {5; - 2} \right)\\
 \Rightarrow \left\{ \begin{array}{l}
3 = b\\
 - 2 = 5a + b
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
b = 3\\
a =  - 1
\end{array} \right.\\
 \Rightarrow BC:y =  - x + 3\\
AH \bot BC\\
 \Rightarrow H\left( {x; - x + 3} \right)\\
 \Rightarrow \overrightarrow {AH}  = \left( {x + 1; - x + 3 - 2} \right) = \left( {x + 1; - x + 1} \right)\\
\overrightarrow {BC}  = \left( {5; - 5} \right)\\
 \Rightarrow \overrightarrow {AH} .\overrightarrow {BC}  = 0\\
 \Rightarrow 5.\left( {x + 1} \right) - 5.\left( { - x + 1} \right) = 0\\
 \Rightarrow x + 1 - \left( { - x + 1} \right) = 0\\
 \Rightarrow x + 1 + x - 1 = 0\\
 \Rightarrow x = 0\\
 \Rightarrow H\left( {0;3} \right)\\
 \Rightarrow H \equiv B\\
 \Rightarrow D\\
2)\\
\left| {3 - x} \right| = \left| {2x - 5} \right|\\
 \Rightarrow \left[ \begin{array}{l}
3 - x = 2x - 5\\
3 - x =  - 2x + 5
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
3x = 8\\
x = 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{8}{3}\\
x = 2
\end{array} \right.\\
 \Rightarrow S = \dfrac{8}{3} + 2 = \dfrac{{14}}{3}\\
 \Rightarrow D\\
3)y = {x^2} - 4x - 1\\
 \Rightarrow I\left( {2; - 5} \right);a = 1 > 0\\
 \Rightarrow B\\
4)\\
\left( {{m^2} - m} \right).x + 2 =  - mx + x + 2m\\
 \Rightarrow \left( {{m^2} - m} \right).x + m.x - x = 2m - 2\\
 \Rightarrow \left( {{m^2} - m + m - 1} \right).x = 2m - 2\\
 \Rightarrow \left( {{m^2} - 1} \right).x = 2\left( {m - 1} \right)\\
 \Rightarrow \left( {m - 1} \right)\left( {m + 1} \right).x = 2\left( {m - 1} \right)\\
 \Rightarrow \left\{ \begin{array}{l}
\left( {m - 1} \right).\left( {m + 1} \right) = 0\\
2\left( {m - 1} \right) = 0
\end{array} \right.\\
 \Rightarrow m = 1\\
 \Rightarrow D
\end{array}$