Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
m \ne 0\\
\Delta ' > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
{\left( {m - 4} \right)^2} - m\left( {m + 8} \right) > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
{m^2} - 8m + 16 - {m^2} - 8m > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
m < 1
\end{array} \right.\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{ - 2m + 8}}{m}\\
{x_1}{x_2} = \frac{{m + 8}}{m}\\
4{x_1} - {x_2} = 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
5{x_1} = \frac{{ - 2m + 8}}{m} + 1\\
{x_1}.\left( {4{x_1} - 1} \right) = \frac{{m + 8}}{m}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} = \frac{{8 - m}}{{5m}}\\
4x_1^2 - {x_1} = \frac{{m + 8}}{m}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x_1} = \frac{{8 - m}}{{5m}}\\
\frac{{\left( {8 - m} \right)\left( {32 - 4m - 5m} \right)}}{{25m}} = m + 8
\end{array} \right.\\
\Rightarrow \left( {8 - m} \right)\left( {32 - 9m} \right) = 25m\left( {m + 8} \right)\\
\Rightarrow 9{m^2} - 72m - 32m + 256 = 25{m^2} + 200m\\
\Rightarrow 16{m^2} + 304m - 256 = 0\\
\Rightarrow m = \frac{{ - 19 \pm 5\sqrt {17} }}{2}\left( {tm} \right)
\end{array}$
