Answer
$(\pm a,0)$, $c=\sqrt{a^2-b^2}$. $(\pm 5,0)$, $(\pm3,0)$
Work Step by Step
The graph of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a\gt b\gt 0$ is an ellipse that has vertices $(\pm a,0)$ and foci $(\pm c,0)$, where $c=\sqrt{a^2-b^2}$.
Hence the graph of $\frac{x^2}{5^2}+\frac{y^2}{4^2}=1$ is an ellipse that has vertices $(\pm 5,0)$ and foci $(\pm c,0)$, where $c=\sqrt{5^2-4^2}=3$.
