Đáp án:

$\begin{array}{l}
a)\left| {5x - 2} \right| \le 13\\
 \Rightarrow  - 13 \le 5x - 2 \le 13\\
 \Rightarrow  - 13 + 2 \le 5x \le 13 + 2\\
 \Rightarrow  - 11 \le 5x \le 15\\
 \Rightarrow  - \frac{{11}}{5} \le x \le \frac{{15}}{5}\\
 \Rightarrow  - 2,2 \le x \le 3\\
 \Rightarrow x \in \left\{ { - 2; - 1;0;1;2;3} \right\}\\
b)\left( {x - 3} \right).\left( {2y + 1} \right) = 5\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 3 = 1\\
2y + 1 = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3 =  - 1\\
2y + 1 =  - 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3 = 5\\
2y + 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3 =  - 5\\
2y + 1 =  - 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 4\\
y = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 2\\
y =  - 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 8\\
y = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  - 2\\
y =  - 1
\end{array} \right.
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {4;2} \right);\left( {2; - 3} \right);\left( {8;0} \right);\left( { - 2; - 1} \right)} \right\}
\end{array}$