Step 1:
$9(x-2)^2-4(y+3)^2=36$
The above equation is divided by $36$ we get
$\large\frac{(x-2)^2}{4}-\large\frac{(y+3)^2}{9}$$=1$
The hyperbola has its centre at $(2,-3)$ and $a^2=4,b^2=9$
$\Rightarrow a=2,b=3$
Step 2:
Let $2\alpha$ be the angle between the asymptotes.
$2\alpha=2\tan^{-1}\large\frac{b}{a}$
$\Rightarrow 2\tan^{-1}\large\frac{3}{2}$
The angle between the asymptotes is $2\tan^{-1}\large\frac{3}{2}$