Answer
$(-a, 0)$. $(a, 0)$. $c=\sqrt {a^2-b^2}$, $(-5, 0)$, $(5, 0)$, $(-3, 0)$, $(3, 0)$
Work Step by Step
1. Recall the definition and the general equation of an ellipse, with the given equation $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$ where $a\gt b\gt0$, we can identify its vertices as $(-a, 0)$ and $(a, 0)$.
2. The foci is given as $(\pm c, 0)$, with the relationship formula $c^2=a^2-b^2$, we have $c=\sqrt {a^2-b^2}$.
3. In the example of $\frac{x^2}{5^2} + \frac{y^2}{4^2}=1$, we have $a=5, b=4$ and $c=\sqrt {5^2-4^2}=3$, so the vertices are $(-5, 0)$ and $(5, 0)$ and the foci are $(-3, 0)$ and $(3, 0)$
