Đáp án:
$\begin{array}{l}
a)m = - 1\\
\Rightarrow {x^2} + 4x - 2 = 0\\
\Rightarrow {x^2} + 4x + 4 = 6\\
\Rightarrow {\left( {x + 2} \right)^2} = 6\\
\Rightarrow \left[ \begin{array}{l}
x = - 2 + \sqrt 6 \\
x = - 2 - \sqrt 6
\end{array} \right.\\
b)\Delta ' \ge 0\\
\Rightarrow {\left( {m - 1} \right)^2} - {m^2} + 3 \ge 0\\
\Rightarrow {m^2} - 2m + 1 - {m^2} + 3 \ge 0\\
\Rightarrow m \le 2\\
THeo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m - 1} \right)\\
{x_1}{x_2} = {m^2} - 3
\end{array} \right.\\
\left| {{x_1} - {x_2}} \right| = 2\\
\Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = 4\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 4\\
\Rightarrow 4{\left( {m - 1} \right)^2} - 4.\left( {{m^2} - 3} \right) = 4\\
\Rightarrow 4{m^2} - 8m + 4 - 4{m^2} + 12 - 4 = 0\\
\Rightarrow - 8m + 12 = 0\\
\Rightarrow m = \dfrac{3}{2}\left( {tmdk} \right)\\
c)\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m - 1} \right)\\
{x_1}{x_2} = {m^2} - 3
\end{array} \right.\\
\Rightarrow m - 1 = \dfrac{{{x_1} + {x_2}}}{2}\\
\Rightarrow m = \dfrac{{{x_1} + {x_2}}}{2} + 1 = \dfrac{{{x_1} + {x_2} + 2}}{2}\\
\Rightarrow {m^2} = \dfrac{{{{\left( {{x_1} + {x_2} + 2} \right)}^2}}}{4}\\
\Rightarrow {x_1}{x_2} + 3 = \dfrac{{{{\left( {{x_1} + {x_2} + 2} \right)}^2}}}{4}\left( { = {m^2}} \right)\\
\Rightarrow {\left( {{x_1} + {x_2} + 2} \right)^2} = 4{x_1}{x_2} + 12
\end{array}$
