$a)f\left(x\right)=\frac{2x+8}{x+3}=\frac{2\left(x+3\right)+2}{x+3}=\frac{2\left(x+3\right)}{x+3}+\frac{2}{x+3}=2+\frac{2}{x+3}=\frac{2}{x+3}+2$

$b)f\left(x\right)=\frac{-3x+10}{x-3}=\frac{-3\left(x-3\right)+1}{x-3}=\frac{-3\left(x-3\right)}{x-3}+\frac{1}{x-3}=-3+\frac{1}{x-3}=\frac{1}{x-3}-3$