Đáp án:
$\begin{array}{l}
a){d_1}:2a\left( {a + 1} \right)x - y = - m - 1\\
\Rightarrow y = \left( {2{a^2} + 2a} \right)x + m + 1\\
{d_2}:4\left( {a - 2} \right).x + y = 3a - 1\\
\Rightarrow y = \left( {8 - 4a} \right).x + 3a - 1\\
{d_1}//{d_2}\\
\Rightarrow \left\{ \begin{array}{l}
2{a^2} + 2a = 8 - 4a\\
m + 1 \ne 3a - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{a^2} + 3a - 4 = 0\\
m \ne 3a - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {a - 1} \right)\left( {a + 4} \right) = 0\\
m \ne 3a - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
a = 1\\
a = - 4
\end{array} \right.\\
m \ne 3a - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 1\\
m \ne - 14
\end{array} \right.\\
b){d_1}:\left( {a + 3} \right).x + \left( {3a - 5} \right).y + a + 7 = 0\\
{d_2}:x - 3y + 11 = 0\\
{d_1} \bot {d_2}\\
\Rightarrow \left( {a + 3} \right).1 + \left( {3a - 5} \right).\left( { - 3} \right) = 0\\
\Rightarrow a + 3 - 9a + 15 = 0\\
\Rightarrow a = \dfrac{9}{4}
\end{array}$
