Đáp án:
a) Để pt có 2 nghiệm phân biệt thì:
$\begin{array}{l}
\left\{ \begin{array}{l}
m + 1 \ne 0\\
\Delta ' > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
m \ne - 1\\
{\left( {m + 2} \right)^2} - \left( {m + 1} \right)\left( {m - 3} \right) > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne - 1\\
6m + 7 > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
m \ne - 1\\
m > - \frac{7}{6}
\end{array} \right.\\
Theo\,viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{2\left( {m + 2} \right)}}{{m + 1}}\\
{x_1}{x_2} = \frac{{m - 3}}{{m + 1}}
\end{array} \right.\\
\left( {4{x_1} + 1} \right)\left( {4{x_2} + 1} \right) - 18 = 0\\
\Rightarrow 4{x_1}{x_2} + 4\left( {{x_1} + {x_2}} \right) + 2 - 18 = 0\\
\Rightarrow \frac{{4m - 12}}{{m + 1}} + \frac{{8m + 16}}{{m + 1}} - 16 = 0\\
\Rightarrow 12m + 4 - 16\left( {m + 1} \right) = 0\\
\Rightarrow m = - 3\left( {ktmdk} \right)\\
\Rightarrow m \in \emptyset \\
b)m \ne - 1;m > - \frac{7}{6}\\
2{x_1} - {x_2} = 0\\
\Rightarrow {x_2} = 2{x_1}\\
\Rightarrow \left\{ \begin{array}{l}
3{x_1} = \frac{{2\left( {m + 2} \right)}}{{m + 1}}\\
2x_1^2 = \frac{{m - 3}}{{m + 1}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x_1} = \frac{{2m + 4}}{{3m + 3}}\\
x_1^2 = \frac{{m - 3}}{{2m + 2}}
\end{array} \right.\\
\Rightarrow {\left( {\frac{{2m + 4}}{{3m + 3}}} \right)^2} = \frac{{m - 3}}{{2m + 2}}\\
\Rightarrow \frac{{4{m^2} + 16m + 16}}{{9\left( {m + 1} \right)}} = m - 3\\
\Rightarrow 4{m^2} + 16m + 16 = 9{m^2} - 18m - 17\\
\Rightarrow 5{m^2} - 34m - 33 = 0\\
\Rightarrow m = \frac{{17 + 6\sqrt {14} }}{5}
\end{array}$
