Answer
The vertices are $(\pm12,0)$.
The foci are $(\pm \sqrt{468},0)$.
Work Step by Step
The equation $$
\left(\frac{x}{2}\right)^{2}-\left(\frac{y}{3}\right)^{2}=36\Longrightarrow\left(\frac{x}{12}\right)^{2}-\left(\frac{y}{18}\right)^{2}=1
$$
is hyperbola. We have $a=12$, $b=18$, $c=\sqrt{a^2+b^2}=\sqrt{468}$ and hence:
The vertices are $(\pm12,0)$.
The foci are $(\pm \sqrt{468},0)$.
