Đáp án:
$\begin{array}{l}
1)Thay\,x = 2\,vao\,pt\,ta\,được:\\
{2^2} - \left( {2m + 1} \right).2 + {m^2} + 2 = 0\\
\Rightarrow {m^2} - 4m + 4 = 0\\
\Rightarrow m = 2\\
Thay\,m = 2\,vao\,pt\,ta\,duoc:\\
{x^2} - 5x + 6 = 0\\
\Rightarrow \left( {x - 2} \right)\left( {x - 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.\\
2)De\,pt\,co\,2\,nghiem:\Delta > 0\\
\Rightarrow {\left( {2m + 1} \right)^2} - 4\left( {{m^2} + 2} \right) > 0\\
\Rightarrow 4{m^2} + 4m + 1 - 4{m^2} - 8 > 0\\
\Rightarrow 4m - 7 > 0\\
\Rightarrow m > \frac{7}{4}\\
Theo\,viet\,ta\,co:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m + 1\\
{x_1}.{x_2} = {m^2} + 2
\end{array} \right.\\
3{x_1}.{x_2} - 5\left( {{x_1} + {x_2}} \right) + 6 = 0\\
\Rightarrow 3\left( {{m^2} + 2} \right) - 5.\left( {2m + 1} \right) + 6 = 0\\
\Rightarrow 3{m^2} + 6 - 10m - 5 + 6 = 0\\
\Rightarrow 3{m^2} - 10m + 7 = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 1\left( {ktm} \right)\\
m = \frac{7}{3}\left( {tm} \right)
\end{array} \right.\\
Vậy\,m = \frac{7}{3}
\end{array}$
