Đáp án:
$\begin{array}{l}
A = \left( {2m - 1;m + 3} \right);\\
B = \left( { - 4;5} \right)\\
a)A \subset B\\
\Rightarrow \left( {2m - 1;m + 3} \right) \subset \left( { - 4;5} \right)\\
\Rightarrow - 4 \le 2m - 1 < m + 3 \le 5\\
\Rightarrow \left\{ \begin{array}{l}
2m - 1 \ge - 4\\
2m - 1 < m + 3\\
m + 3 \le 5
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ge - \dfrac{3}{2}\\
m < 4\\
m \le 2
\end{array} \right.\\
\Rightarrow - \dfrac{3}{2} \le m \le 2\\
b)B \subset A\\
\Rightarrow \left( { - 4;5} \right) \subset \left( {2m - 1;m + 3} \right)\\
\Rightarrow 2m - 1 \le - 4 < 5 \le m + 3\\
\Rightarrow \left\{ \begin{array}{l}
2m - 1 \le - 4\\
5 \le m + 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \le - \dfrac{3}{2}\\
m \ge 2
\end{array} \right.\left( {ktm} \right)\\
\Rightarrow m \in \emptyset \\
c)A \cap B = \emptyset \\
\Rightarrow \left[ \begin{array}{l}
m + 3 < - 4\\
5 < 2m - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
m < - 7\\
m > 3
\end{array} \right.\\
Vay\,m < - 7\,hoac\,m > 3\\
d)\exists A \cap B\\
\Rightarrow \left[ \begin{array}{l}
2m - 1 \le - 4 < m + 3 \le 5\\
- 4 \le 2m - 1 < 5 \le m + 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m \le - \dfrac{3}{2}\\
m > - 7\\
m \le 2
\end{array} \right.\\
\left\{ \begin{array}{l}
m \ge - \dfrac{3}{2}\\
m < 3\\
m \ge 2
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
- 7 < m \le \dfrac{3}{2}\\
2 \le m < 3
\end{array} \right.
\end{array}$
