$a)\text{ĐKXĐ}:\left\{\begin{array}{l} a+3 \ne 0\\ a+2 \ne 0 \\ a-2 \ne 0 \\ 4-a^2 \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} a \ne -3\\ a \ne \pm 2 \end{array} \right.\\ b)A=\dfrac{2a-a^2}{a+3}:\left(\dfrac{a-2}{a+2}-\dfrac{a+2}{a-2}+\dfrac{4a^2}{4-a^2}\right)\\ =\dfrac{a(2-a)}{a+3}:\left(\dfrac{(a-2)^2}{(a-2)(a+2)}-\dfrac{(a+2)^2}{(a-2)(a+2)}-\dfrac{4a^2}{(a-2)(a+2)}\right)\\ =\dfrac{a(2-a)}{a+3}:\dfrac{(a-2)^2-(a+2)^2-4a^2}{(a-2)(a+2)}\\ =\dfrac{a(2-a)}{a+3}:\dfrac{-8a-4a^2}{(a-2)(a+2)}\\ =\dfrac{a(2-a)}{a+3}.\dfrac{(a-2)(a+2)}{-4a(a+2)}\\ =\dfrac{(2-a)^2}{4(a+3)}$
