Đáp án:
a) Thay m=-1 vào pt ta được:
$\begin{array}{l}
{x^2} + \left( { - 1 - 1} \right)x + 5.\left( { - 1} \right) - 6 = 0\\
\Rightarrow {x^2} - 2x - 11 = 0\\
\Rightarrow {x^2} - 2x + 1 = 12\\
\Rightarrow {\left( {x - 1} \right)^2} = 12\\
\Rightarrow \left[ \begin{array}{l}
x - 1 = 2\sqrt 3 \\
x - 1 = - 2\sqrt 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1 + 2\sqrt 3 \\
x = 1 - 2\sqrt 3
\end{array} \right.\\
b)\\
{x^2} + \left( {m - 1} \right)x + 5m - 6 = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow {\left( {m - 1} \right)^2} - 4\left( {5m - 6} \right) > 0\\
\Rightarrow {m^2} - 2m + 1 - 20m + 24 > 0\\
\Rightarrow {m^2} - 22m + 25 > 0\\
\Rightarrow \left[ \begin{array}{l}
m > 11 + 4\sqrt 6 \\
m < 11 - 4\sqrt 6
\end{array} \right.\\
Theo\,viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 1 - m\\
{x_1}{x_2} = 5m - 6\\
4{x_1} + 3{x_2} = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
3{x_1} + 3{x_2} = 3 - 3m\\
4{x_1} + 3{x_2} = 1
\end{array} \right.\\
\Rightarrow {x_1} = 3m - 2\\
\Rightarrow {x_2} = 1 - m - \left( {3m - 2} \right) = 3 - 4m\\
\Rightarrow {x_1}.{x_2} = 5m - 6\\
\Rightarrow \left( {3m - 2} \right).\left( {3 - 4m} \right) = 5m - 6\\
\Rightarrow - 12{m^2} + 9m + 8m - 6 = 5m - 6\\
\Rightarrow 12{m^2} - 12m = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 0\left( {tm} \right)\\
m = 1\left( {tm} \right)
\end{array} \right.\\
Vậy\,m = 0/m = 1
\end{array}$
