Đáp án:
$\begin{array}{l}
1)\Delta > 0\\
\Rightarrow {\left( {m + 2} \right)^2} - 4\left( { - m - 4} \right) > 0\\
\Rightarrow {m^2} + 4m + 4 + 4m + 16 > 0\\
\Rightarrow {m^2} + 8m + 16 + 4 > 0\\
\Rightarrow {\left( {m + 4} \right)^2} + 4 > 0\left( {tm} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - m - 2\\
{x_1}{x_2} = - m - 4
\end{array} \right.\\
{x_1} < 0 \le {x_2}\\
\Rightarrow {x_1}.{x_2} \le 0\\
\Rightarrow - m - 4 \le 0\\
\Rightarrow m \ge - 4\\
2)\\
\dfrac{1}{2}{x^2} = \left( {m - 1} \right)x + m\\
\Rightarrow {x^2} - 2\left( {m - 1} \right)x - 2m = 0\\
\Rightarrow \Delta ' > 0\\
\Rightarrow {\left( {m - 1} \right)^2} + 2m > 0\\
\Rightarrow {m^2} + 1 > 0\left( {tm} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m - 1} \right)\\
{x_1}{x_2} = - 2m
\end{array} \right.\\
a){x_1} < 2 < {x_2}\\
\Rightarrow \left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right) < 0\\
\Rightarrow {x_1}{x_2} - 2\left( {{x_1} + {x_2}} \right) + 4 < 0\\
\Rightarrow - 2m - 2.2\left( {m - 1} \right) + 4 < 0\\
\Rightarrow - 2m - 4m + 4 + 4 < 0\\
\Rightarrow m > \dfrac{4}{3}\\
b){x_1} < {x_2} < 2\\
\Rightarrow \left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right) > 0\\
\Rightarrow m < \dfrac{4}{3}\\
c){x_1} > {x_2} > 2\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} > 4\\
\left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right) > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2\left( {m - 1} \right) > 4\\
m < \dfrac{4}{3}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > 3\\
m < \dfrac{4}{3}
\end{array} \right.\\
\Rightarrow m \in \emptyset
\end{array}$
