Đáp án:
$\begin{array}{l}
e)\left( 1 \right) \equiv y = 3x - 4\\
\Leftrightarrow \left\{ \begin{array}{l}
m - 1 = 3\\
m - 2 = - 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m = 4\\
m = - 2
\end{array} \right.\left( {ktm} \right)\\
Vậy\,m \in \emptyset \\
f)Khi:x = 1\\
\Leftrightarrow y = 2x - 3 = 2.1 - 3 = - 1\\
\Leftrightarrow \left( {1; - 1} \right) \in \left( 1 \right)\\
\Leftrightarrow - 1 = \left( {m - 1} \right).1 + m - 2\\
\Leftrightarrow 2m - 3 = - 1\\
\Leftrightarrow 2m = 2\\
\Leftrightarrow m = 1\\
Vậy\,m = 1\\
g)Khi:y = 2\\
\Leftrightarrow y = 5 - 3x = 2\\
\Leftrightarrow 3x = 3\\
\Leftrightarrow x = 1\\
\Leftrightarrow \left( {1;2} \right) \in \left( 1 \right)\\
\Leftrightarrow 2 = \left( {m - 1} \right).1 + m - 2\\
\Leftrightarrow 2m - 3 = 2\\
\Leftrightarrow 2m = 5\\
\Leftrightarrow m = \dfrac{5}{2}\\
Vậy\,m = \dfrac{5}{2}\\
h)y = 3x - 2\\
Khi:x = 0 \Leftrightarrow y = - 2\\
\Leftrightarrow \left( {0; - 2} \right) \in \left( 1 \right)\\
\Leftrightarrow - 2 = \left( {m - 1} \right).0 + m - 2\\
\Leftrightarrow m = 0\\
Vậy\,m = 0
\end{array}$
