Answer
Foci: $(\pm 10,0)$,
Vertices: $(\pm 6,0)$,
Asymptotes: $y=\pm\frac{4}{3}x$.
Work Step by Step
Recall: The hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ has the foci $(\pm c,0)$ where $c^2=a^2+b^2$, the vertices $(\pm a,0)$, and the asymptotes $y=\pm(b/a)x$
We have $\frac{x^2}{36}-\frac{y^2}{64}=1$.
Then,
$a^2=36\to a=6$
$b^2=64\to b=8$
$c=\sqrt{a^2+b^2}=\sqrt{36+64}=\sqrt{100}=10$
So, the foci are $(\pm 10,0)$, the vertices are $(\pm 6,0)$, and the asymptotes are $y=\pm\frac{8}{6}x=\pm\frac{4}{3}x$.
