Đáp án:
$\begin{array}{l}
{x^2} - 2\left( {m + 1} \right)x + 4{m^2} - 4m - 1 = 0\\
\Rightarrow \Delta ' > 0\\
\Rightarrow {\left( {m + 1} \right)^2} - 4{m^2} + 4m + 1 > 0\\
\Rightarrow {m^2} + 2m + 1 - 4{m^2} + 4m + 1 > 0\\
\Rightarrow 3{m^2} - 6m - 2 < 0\\
\Rightarrow \dfrac{{3 - \sqrt {15} }}{3} < m < \dfrac{{3 + \sqrt {15} }}{3}\\
TheoViet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 1} \right)\\
{x_1}{x_2} = 4{m^2} - 4m - 1
\end{array} \right.\\
\left| {{x_1} - 2} \right| > \left| {{x_2} + 1} \right|\\
\Rightarrow \left[ \begin{array}{l}
{x_1} - 2 > {x_2} + 1\\
{x_1} - 2 < - {x_2} - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x_1} - {x_2} > 3\\
{x_1} + {x_2} < 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} > 9\\
2\left( {m + 1} \right) < 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
4{\left( {m + 1} \right)^2} - 4.\left( {4{m^2} - 4m - 1} \right) > 9\left( * \right)\\
m < - \dfrac{1}{2}
\end{array} \right.\\
\left( * \right) \Rightarrow 4{m^2} + 8m + 4 - 16{m^2} + 16m + 4 - 9 > 0\\
\Rightarrow - 12{m^2} + 24m - 1 > 0\\
\Rightarrow 12{m^2} - 24m + 1 < 0\\
\Rightarrow \dfrac{{6 - \sqrt {33} }}{6} < m < \dfrac{{6 + \sqrt {33} }}{6}\\
Vậy\,\dfrac{{6 - \sqrt {33} }}{6} < m < \dfrac{{6 + \sqrt {33} }}{6}
\end{array}$
