Đáp án:
$\begin{array}{l}
a)\left\{ \begin{array}{l}
x + 4{m^2} \le 2mx + 1\\
3x + 2 > 2x - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x - 2mx \le 1 - 4{m^2}\\
3x - 2x > - 1 - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {1 - 2m} \right).x \le \left( {1 - 2m} \right)\left( {1 + 2m} \right)\\
x > - 3
\end{array} \right.\\
+ Khi:m = \dfrac{1}{2} \Rightarrow 0.x \le 0\left( {tm} \right)\\
+ Khi:m < \dfrac{1}{2} \Rightarrow \left\{ \begin{array}{l}
x \le 1 + 2m\\
x > - 3
\end{array} \right.\\
\Rightarrow 1 + 2m > - 3\\
\Rightarrow 2m > - 4\\
\Rightarrow m > - 2\\
\Rightarrow - 2 < m < \dfrac{1}{2}\\
+ Khi:m > \dfrac{1}{2} \Rightarrow \left\{ \begin{array}{l}
x \ge 1 + 2m\\
x > - 3
\end{array} \right.\left( {tm} \right)\\
Vậy\,m > - 2\\
b)\left\{ \begin{array}{l}
4x - 5 > - 3x + 2\\
3x + 2m + 2 < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
7x > 7\\
3x < - 2m - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x > 1\\
x < \dfrac{{ - 2m - 2}}{3}
\end{array} \right.\\
\Rightarrow 1 < \dfrac{{ - 2m - 2}}{3}\\
\Rightarrow - 2m - 2 > 3\\
\Rightarrow 2m < - 5\\
\Rightarrow m < - \dfrac{5}{2}\\
Vậy\,m < - \dfrac{5}{2}
\end{array}$
