Đáp án:
$\begin{array}{l}
a)m = 4\\
\Leftrightarrow {x^2} - 2x - 3 = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow x = 3;x = - 1\\
Vậy\,x = 3;x = - 1\,khi:m = 4\\
b)\Delta > 0\\
\Leftrightarrow {\left( {m - 2} \right)^2} - 4.\left( { - 3} \right) > 0\\
\Leftrightarrow {\left( {m - 2} \right)^2} + 12 > 0\left( {tmdk} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m - 2\\
{x_1}{x_2} = - 3
\end{array} \right.\\
\sqrt {x_1^2 + 2020} - {x_1} = \sqrt {x_2^2 + 2020} + {x_2}\\
\Leftrightarrow \sqrt {x_1^2 + 2020} - \sqrt {x_2^2 + 2020} = {x_1} + {x_2} = m - 2\\
\Leftrightarrow x_1^2 + 2020 - 2\sqrt {x_1^2 + 2020} .\sqrt {x_2^2 + 2020} + x_2^2 + 2020\\
= {\left( {m - 2} \right)^2}\\
\Leftrightarrow x_1^2 + x_2^2 + 4040 - 2.\sqrt {x_1^2x_2^2 + 2020\left( {x_1^2 + x_2^2} \right) + {{2020}^2}} \\
= {\left( {m - 2} \right)^2}\left( 1 \right)\\
Xet:x_1^2 + x_2^2\\
= {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2}\\
= {\left( {m - 2} \right)^2} - 2.\left( { - 3} \right)\\
= {m^2} - 4m + 10\\
\left( 1 \right) \Leftrightarrow {m^2} - 4m + 10 + 4040\\
- 2.\sqrt {{{\left( { - 3} \right)}^2} + 2020.\left( {{m^2} - 4m + 10} \right) + {{2020}^2}} \\
= {m^2} - 4m + 4\\
\Leftrightarrow 4046 - 2\sqrt {2020\left( {{m^2} - 4m + 10} \right) + 4080409} = 0\\
\Leftrightarrow \sqrt {2020\left( {{m^2} - 4m + 10} \right) + 4080409} = 2023\\
\Leftrightarrow 2020\left( {{m^2} - 4m + 10} \right) + 4080409 = 4092529\\
\Leftrightarrow 2020\left( {{m^2} - 4m + 10} \right) = 12120\\
\Leftrightarrow {m^2} - 4m + 10 = 6\\
\Leftrightarrow {m^2} - 4m + 4 = 0\\
\Leftrightarrow {\left( {m - 2} \right)^2} = 0\\
\Leftrightarrow m = 2\\
Vậy\,m = 2
\end{array}$