Đáp án:
$\begin{array}{l}
a)\left( {14;2020} \right) \in \left( d \right)\\
\Leftrightarrow 2020 = 2.14 - m + 1\\
\Leftrightarrow m = 28 + 1 - 2020\\
\Leftrightarrow m = - 1991\\
Vay\,m = - 1991\\
b)\\
Xét:\dfrac{1}{2}{x^2} = 2x - m + 1\\
\Leftrightarrow {x^2} - 4x + 2m - 2 = 0\\
\Delta ' = 4 - 2m + 2 = 6 - 2m\\
\Delta ' > 0\\
\Leftrightarrow 6 - 2m > 0\\
\Leftrightarrow m < 3\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 4\\
{x_1}{x_2} = 2m - 2
\end{array} \right.\\
{y_1} = 2{x_1} - m + 1\\
{y_2} = 2{x_2} - m + 1\\
Khi:{x_1}{x_2}\left( {{y_1} + {y_2}} \right) + 48 = 0\\
\Leftrightarrow \left( {2m - 2} \right).\left( {2{x_1} - m + 1 + 2{x_2} - m + 1} \right) + 48 = 0\\
\Leftrightarrow 2\left( {m - 1} \right).\left( {2\left( {{x_1} + {x_2}} \right) - 2m + 2} \right) + 48 = 0\\
\Leftrightarrow \left( {m - 1} \right)\left( {{x_1} + {x_2} - m + 1} \right) + 12 = 0\\
\Leftrightarrow \left( {m - 1} \right).\left( {4 - m + 1} \right) + 12 = 0\\
\Leftrightarrow \left( {m - 1} \right)\left( {5 - m} \right) + 12 = 0\\
\Leftrightarrow - {m^2} + 6m - 5 + 12 = 0\\
\Leftrightarrow {m^2} - 6m - 7 = 0\\
\Leftrightarrow \left( {m - 7} \right)\left( {m + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
m = 7\left( {ktm} \right)\\
m = - 1\left( {tm} \right)
\end{array} \right.\\
Vậy\,m = - 1
\end{array}$
