Đáp án:

$\begin{array}{l}
1)\left( {d'} \right):y = 2x - 3\\
Khi:y = 1\\
 \Leftrightarrow 2x - 3 = 1\\
 \Leftrightarrow 2x = 4\\
 \Leftrightarrow x = 2\\
 \Leftrightarrow A\left( { - 1;2} \right);\left( {2;1} \right) \in \left( d \right)\\
 \Leftrightarrow \left\{ \begin{array}{l}
2 =  - a + b\\
1 = 2a + b
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
3a =  - 1\\
b = 1 - 2a
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
a =  - \dfrac{1}{3}\\
b = 1 - 2.\dfrac{{ - 1}}{3} = \dfrac{5}{3}
\end{array} \right.\\
Vậy\,a = \dfrac{{ - 1}}{3};b = \dfrac{5}{3}\\
2)\Delta  > 0\\
 \Leftrightarrow {\left( {m - 1} \right)^2} - 4.\left( { - 2} \right) > 0\\
 \Leftrightarrow {\left( {m - 1} \right)^2} + 8 > 0\left( {tm} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m - 1\\
{x_1}{x_2} =  - 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x_1^2 - \left( {m - 1} \right){x_1} - 2 = 0\\
x_2^2 - \left( {m - 1} \right){x_2} - 2 = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x_1^2 = \left( {m - 1} \right){x_1} + 2\\
x_2^2 = \left( {m - 1} \right){x_2} + 2
\end{array} \right.\\
Khi:\dfrac{{x_1^2 + 2{x_1} - m + 1}}{{{x_2} + 3}}:\dfrac{{{x_1} + 3}}{{x_2^2 + 2{x_2} - m + 1}} = 1\\
 \Leftrightarrow \dfrac{{\left( {m - 1} \right){x_1} + 2 + 2{x_1} - m + 1}}{{{x_2} + 3}}.\dfrac{{\left( {m - 1} \right){x_2} + 2 + 2{x_2} - m + 1}}{{{x_1} + 3}} = 1\\
 \Leftrightarrow \left[ {\left( {m + 1} \right).{x_1} - m + 3} \right].\left[ {\left( {m + 1} \right).{x_2} - m + 3} \right]\\
 = \left( {{x_2} + 3} \right)\left( {{x_1} + 3} \right)\\
 \Leftrightarrow {\left( {m + 1} \right)^2}.{x_1}{x_2} + \left( { - m + 3} \right).\left( {m + 1} \right).\left( {{x_1} + {x_2}} \right) + {\left( { - m + 3} \right)^2}\\
 = {x_1}{x_2} + 3\left( {{x_1} + {x_2}} \right) + 9\\
 \Leftrightarrow  - 2{\left( {m + 1} \right)^2} + \left( { - m + 3} \right).\left( {m + 1} \right)\left( {m - 1} \right) + {\left( {m - 3} \right)^2}\\
 =  - 2 + 3.\left( {m - 1} \right) + 9\\
 \Leftrightarrow  - 2{m^2} - 4m - 2 + \left( { - m + 3} \right).\left( {{m^2} - 1} \right) + {m^2} - 6m + 9\\
 =  - 2 + 3m - 3 + 9\\
 \Leftrightarrow  - {m^2} - 10m + 7 - 3m - 4 - {m^3} - m + 3{m^2} - 3 = 0\\
 \Leftrightarrow  - {m^3} + 2{m^2} - 14m = 0\\
 \Leftrightarrow  - m\left( {{m^2} - 2m + 14} \right) = 0\\
 \Leftrightarrow m = 0\left( {do:{m^2} - 2m + 14 > 0} \right)\\
Khi:m = 0\\
 \Leftrightarrow {x^2} + x - 2 = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow x = 1;x =  - 2\\
Vậy\,x = 1;x =  - 2\,khi:m = 0
\end{array}$