Đáp án:
$\begin{array}{l}
a)\Delta ' = {\left( {m - 3} \right)^2} - \left( { - 4 - {m^2}} \right)\\
= {\left( {m - 3} \right)^2} + {m^2} + 4 > 0
\end{array}$
Vậy pt luôn có 2 nghiệm phân biệt.
$\begin{array}{l}
b)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m - 3} \right)\\
{x_1}{x_2} = - 4 - {m^2}
\end{array} \right.\\
\left| {{x_1}} \right| - \left| {{x_2}} \right| = 6\\
\Leftrightarrow x_1^2 - 2\left| {{x_1}{x_2}} \right| + x_2^2 = 36\\
\Leftrightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} - 2\left| {{x_1}{x_2}} \right| = 36\\
\Leftrightarrow 4{\left( {m - 3} \right)^2} - 2.\left( { - 4 - {m^2}} \right) - 2\left| { - 4 - {m^2}} \right| = 36\\
\Leftrightarrow 4{\left( {m - 3} \right)^2} = 36\\
\Leftrightarrow {\left( {m - 3} \right)^2} = 9\\
\Leftrightarrow \left[ \begin{array}{l}
m - 3 = 3\\
m - 3 = - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
m = 6\\
m = 0
\end{array} \right.\\
Vậy\,m = 0;m = 6
\end{array}$
