$\begin{array}{l}a)\,\,(x+3)(2-x)>0\\\text{$\to x+3;2-x$ cùng dấu}\\\to\left[ \begin{array}{l}\begin{cases} x+3>0\\2-x>0\end{cases}\\\begin{cases} x+3<0\\2-x<0\end{cases}\end{array} \right.\\\to\left[ \begin{array}{l}\begin{cases} x>-3\\-x>-2\end{cases}\\\begin{cases} x<-3\\-x<-2\end{cases}\end{array} \right.\\\to\left[ \begin{array}{l}\begin{cases} x>-3\\x<2\end{cases} \,\,\rm (TM)\\\begin{cases} x<-3\\x>2\end{cases} \,\,\rm (KTM)\end{array} \right. \\\to x\in\{-2;-1;0;1\}\\\,\\b)\,\,a+b=ab\\\Leftrightarrow ab-(a+b)=0\\\Leftrightarrow ab-a-b=0\\\Leftrightarrow a(b-1)-b=0\\\Leftrightarrow a(b-1)-(b-1)=0+1\\\Leftrightarrow (a-1)(b-1)=1\\\text{mà $a,b\in\mathbb{Z}$}\\\to a-1,b-1\in Ư(1)=\{\pm1\}\\\text{- Ta có bảng sau :}\\\begin{array}{|c|c|}\hline a-1&-1&1\\\hline b-1&-1&1\\\hline a&2&0\\\hline b&2&0\\\hline \end{array}\\\text{- Vậy các cặp số $(x,y)$ thỏa mãn là : $(2,2);(0,0)$}\end{array}$
