Đáp án:
$\begin{array}{l}
a){x^2} - 2x - 3{m^2} + m - 2 = 0\\
m = - 1\\
\Leftrightarrow {x^2} - 2x - 3 - 1 - 2 = 0\\
\Leftrightarrow {x^2} - 2x - 6 = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2} = 7\\
\Leftrightarrow x = 1 \pm \sqrt 7 \\
b)\Delta ' = 1 - \left( { - 3{m^2} + m - 2} \right)\\
= 3{m^2} - m + 3\\
= 3.\left( {{m^2} - \dfrac{1}{3}m + 1} \right)\\
= 3.\left( {{m^2} - 2.m.\dfrac{1}{6} + \dfrac{1}{{36}} + \dfrac{{35}}{{36}}} \right)\\
= 3.{\left( {m - \dfrac{1}{6}} \right)^2} + \dfrac{{35}}{{12}} > 0
\end{array}$
Vậy pt luôn có 2 nghiệm với mọi m
$\begin{array}{l}
c)x = 2\\
\Leftrightarrow {2^2} - 2.2 - 3{m^2} + m - 2 = 0\\
\Leftrightarrow 3{m^2} - m + 2 = 0\left( {vn} \right)
\end{array}$
Vậy ko có m để pt có nghiệm x=2
$\begin{array}{l}
d)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = - 3{m^2} + m - 2
\end{array} \right.\\
Do:x_1^2 - 2{x_1} - 3{m^2} + m - 2 = 0\\
\Leftrightarrow x_1^2 = 2{x_1} + 3{m^2} - m + 2\\
x_1^2 - 2{x_2} + {x_1}{x_2} = 8\\
\Leftrightarrow 2{x_1} + 3{m^2} - m + 2 - 2{x_2} + {x_1}{x_2} = 8\\
\Leftrightarrow 2{x_1} - 2{x_2} = 8\\
\Leftrightarrow {x_1} - {x_2} = 4\\
\Leftrightarrow \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2}} = 4\\
\Leftrightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 16\\
\Leftrightarrow 4 - 4.\left( { - 3{m^2} + m - 2} \right) = 16\\
\Leftrightarrow 1 + 3{m^2} - m + 2 = 4\\
\Leftrightarrow 3{m^2} - m - 1 = 0\\
\Leftrightarrow m = \dfrac{{1 \pm \sqrt {13} }}{6}
\end{array}$
