$\begin{array}{l}
a)A = \left\{ {x \in \mathbb{Z}|2{x^2} - 5x + 3 = 0} \right\}\\
\Rightarrow 2{x^2} - 5x + 3 = 0 \Leftrightarrow \left( {2x - 3} \right)\left( {x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = 1
\end{array} \right. \Rightarrow A = \left\{ 1 \right\}\left( {do\,x \in \mathbb{Z}} \right)\\
b)B = \left\{ {x \in {\mathbb{N}^*}|x \le 16,x \vdots 4} \right\}\\
\Rightarrow B = \left\{ {4;8;12;16} \right\}\\
c)C = \left\{ {x \in \mathbb{Q}|\left( {{x^2} - 2} \right)\left( {2x + 1} \right) = 0} \right\}\\
\left( {{x^2} - 2} \right)\left( {2x + 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 2(L)\\
x = - \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow C = \left\{ { - \dfrac{1}{2}} \right\}\\
d)D = \left\{ {x \in \mathbb{Z}|\left| {4x + 1} \right| = 9} \right\}\\
\left| {4x + 1} \right| = 9 \Leftrightarrow \left[ \begin{array}{l}
4x + 1 = 9\\
4x + 1 = - 9
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
4x = 8\\
4x = - 10
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = - \dfrac{5}{2}(L)
\end{array} \right.\\
\Rightarrow A = \left\{ 2 \right\}
\end{array}$
