Đáp án:
1) Để pt có 2 nghiệm phân biệt thì
$\begin{array}{l}
\Delta ' > 0\\
\Leftrightarrow 4 - m + 2 > 0\\
\Leftrightarrow m < 6\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 4\\
{x_1}{x_2} = m - 2
\end{array} \right.\\
x_1^2 - 2{x_1}{x_2} + 2{x_2} = 1\\
\Leftrightarrow x_1^2 - 2.{x_1}\left( {4 - {x_1}} \right) + 2\left( {4 - {x_1}} \right) = 1\\
\left( {do:{x_2} = 4 - {x_1}} \right)\\
\Leftrightarrow x_1^2 - 8{x_1} + 2x_1^2 + 8 - 2{x_1} - 1 = 0\\
\Leftrightarrow 3x_1^2 - 10{x_1} + 7 = 0\\
\Leftrightarrow \left( {3{x_1} - 7} \right)\left( {{x_1} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x_1} = \dfrac{7}{3};{x_2} = 4 - {x_1} = \dfrac{5}{3}\\
{x_1} = 1;{x_2} = 4 - {x_1} = 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
m - 2 = \dfrac{7}{3}.\dfrac{5}{3} \Leftrightarrow m = \dfrac{{53}}{9}\left( {tm} \right)\\
m - 2 = 1.3 \Leftrightarrow m = 5\left( {tm} \right)
\end{array} \right.\\
Vậy\,m = \dfrac{{53}}{9};m = 5\\
2)Khi:x = 1\\
\Leftrightarrow y = {x^2} = {1^2} = 1\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {1;1} \right) \in y = a.x + b\\
y = a.x + b//y = - x + 5
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
1 = a + b\\
a = - 1\\
b\# 5
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 1 - a = 2\\
a = - 1\\
b\# 5
\end{array} \right.\\
\Leftrightarrow a = - 1;b = 2\\
Vậy\,a = - 1;b = 2
\end{array}$
