$a)$

$\{\begin{array}{l}x-3\ne 0\\ x+3\ne 0\end{array}\Rightarrow \{\begin{array}{l}x\ne 3\\ x\ne -3\end{array}\Rightarrow D=\mathbb{R}\backslash \left\{-3,3\right\}$

$\frac{2}{x-3}+\frac{4}{x+3}=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2x+6}{x^2-9}+\frac{4x-12}{x^2-9}=\frac{2x+6+4x-12}{x^2-9}=\frac{6x-6}{x^2-9}$

$b)$