$\begin{array}{l}
y = \frac{{{x^3}}}{3} - \left( {m - 2} \right){x^2} + \left( {4m - 8} \right)x + m + 1\\
\Rightarrow y' = {x^2} - 2\left( {m - 2} \right)x + 4m - 8\\
\Rightarrow y' = 0\\
\Leftrightarrow {x^2} - 2\left( {m - 2} \right)x + 4m - 8 = 0\\
co:\,\,\,\Delta = {\left( {m - 2} \right)^2} - 4m + 8 = {m^2} - 8m + 12\\
\Rightarrow hs\,co\,\,2\,\,cuc\,\,tri\,\, \Leftrightarrow \Delta > 0\\
\Leftrightarrow {m^2} - 8m + 12 > 0 \Leftrightarrow \left[ \begin{array}{l}
m > 6\\
m < 2
\end{array} \right..\\
Ap\,\,dung\,\,\,dinh\,\,ly\,\,vi - et\,\,ta\,\,co:\,\,\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m - 4\\
{x_1}{x_2} = 4m - 8
\end{array} \right..\\
{x_1} < - 2 < {x_2}\\
\Leftrightarrow \left( {{x_1} + 2} \right)\left( {{x_2} + 2} \right) < 0\\
\Leftrightarrow {x_1}{x_2} + 2\left( {{x_1} + {x_2}} \right) + 4 < 0\\
\Leftrightarrow 4m - 8 + 2\left( {2m - 4} \right) + 4 < 0\\
\Leftrightarrow 4m - 8 + 4m - 8 + 4 < 0\\
\Leftrightarrow 8m < 12\\
\Leftrightarrow m < \frac{3}{5}.\\
Doi\,\,chieu\,\,voi\,\,dk \Rightarrow m < \frac{3}{2}\,\,\,thoa\,man\,\,\,bai\,\,toan.
\end{array}$
