Đáp án:
$\begin{array}{l}
a)n = 0\\
\Leftrightarrow {x^2} - 2mx + 2m - 1 = 0\\
\Delta ' = {m^2} - 2m + 1 = {\left( {m - 1} \right)^2} \ge 0
\end{array}$
Vậy pt luôn có nghiệm khi $n = 0$
$\begin{array}{l}
\left\{ \begin{array}{l}
{x_1} + {x_2} = - 1\\
x_1^2 + x_2^2 = 13
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = - 1\\
{\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 13
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = - 1\\
1 - 2{x_1}{x_2} = 13
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = - 1\\
{x_1}{x_2} = - 6
\end{array} \right.\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - 1 = 2m - n\\
{x_1}{x_2} = - 6 = 2m + 3n - 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2m - n = - 1\\
2m + 3n = - 5
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
4n = - 4\\
2m = n - 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
n = - 1\\
2m = - 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
n = - 1\\
m = - 1
\end{array} \right.\\
Vậy\,m = n = - 1
\end{array}$
