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