Đáp án:

\[\begin{array}{l}
a,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x;y} \right) \in \left\{ {\left( { - 1;4} \right);\left( { - 3; - 2} \right);\left( {1;2} \right);\left( { - 5;0} \right)} \right\}\\
b,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x;y} \right) \in \left\{ {\left( {2; - 3} \right);\left( {4; - 1} \right)} \right\}
\end{array}\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
\left( {x + 2} \right)\left( {y - 1} \right) = 3\\
x,y \in Z \Rightarrow \left( {x + 2} \right);\left( {y - 1} \right) \in Z\\
3 = 1.3 = \left( { - 1} \right).\left( { - 3} \right)\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 = 1\\
y - 1 = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 =  - 1\\
y - 1 =  - 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 = 3\\
y - 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 =  - 3\\
y - 1 =  - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x =  - 1\\
y = 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  - 3\\
y =  - 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1\\
y = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  - 5\\
y = 0
\end{array} \right.
\end{array} \right.\\
 \Rightarrow \left( {x;y} \right) \in \left\{ {\left( { - 1;4} \right);\left( { - 3; - 2} \right);\left( {1;2} \right);\left( { - 5;0} \right)} \right\}\\
b,\\
\left( {3 - x} \right)\left( {xy + 5} \right) =  - 1\\
x;y \in Z \Rightarrow \left( {3 - x} \right);\left( {xy + 5} \right) \in Z\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3 - x = 1\\
xy + 5 =  - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
3 - x =  - 1\\
xy + 5 = 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 2\\
xy =  - 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 4\\
xy =  - 4
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 2\\
y =  - 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 4\\
y =  - 1
\end{array} \right.
\end{array} \right.\\
 \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {2; - 3} \right);\left( {4; - 1} \right)} \right\}
\end{array}\)