Đáp án:
$\begin{array}{l}
1)M\left( { - 2;3} \right) \in \left( d \right)\\
\Leftrightarrow 3 = 2.\left( { - 2} \right) + m\\
\Leftrightarrow m = 3 + 4\\
\Leftrightarrow m = 7\\
Vậy\,m = 7\\
2)Xet:2{x^2} = 2x + m\\
\Leftrightarrow 2{x^2} - 2x - m = 0\\
\Leftrightarrow \Delta ' > 0\\
\Leftrightarrow 1 - 2.\left( { - m} \right) > 0\\
\Leftrightarrow 2m > - 1\\
\Leftrightarrow m > \dfrac{{ - 1}}{2}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 1\\
{x_1}{x_2} = \dfrac{{ - m}}{2}
\end{array} \right.\\
{y_1} = 2{x_1} + m;{y_2} = 2{x_2} + m\\
Khi:{\left( {1 - {x_1}{x_2}} \right)^2} + 2\left( {{y_1} + {y_2}} \right) = 16\\
\Leftrightarrow {\left( {1 - \dfrac{{ - m}}{2}} \right)^2} + 2.\left( {2{x_1} + m + 2{x_2} + m} \right) = 16\\
\Leftrightarrow \dfrac{{{{\left( {m + 2} \right)}^2}}}{4} + 2.\left( {2.\left( {{x_1} + {x_2}} \right) + 2m} \right) = 16\\
\Leftrightarrow \dfrac{{{{\left( {m + 2} \right)}^2}}}{4} + 2.\left( {2 + 2m} \right) = 16\\
\Leftrightarrow \dfrac{{{{\left( {m + 2} \right)}^2}}}{4} + 4m - 12 = 0\\
\Leftrightarrow {\left( {m + 2} \right)^2} + 16m - 48 = 0\\
\Leftrightarrow {m^2} + 20m - 44 = 0\\
\Leftrightarrow \left( {m - 2} \right)\left( {m + 22} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
m = 2\left( {tm} \right)\\
m = - 22\left( {ktm} \right)
\end{array} \right.\\
Vậy\,m = 2
\end{array}$
