Answer
The vertices are $(3,-1\pm7)$ and $(3\pm 4,-1)$ .
The foci are $(3,-1\pm \sqrt{33})$.
Work Step by Step
The equation $$
\left(\frac{x-3}{4}\right)^{2}+\left(\frac{y+1}{7}\right)^{2}=1
$$
is an ellipse with center $(3,-1)$. We have $a=4$, $b=7$, $c=\sqrt{b^2-a^2}=\sqrt{33}$ and hence:
When the center is at the origin, the foci are $(0,\pm \sqrt{33})$ and the vertices are $(\pm 4,0)$ and $(0.\pm 7)$ Then the translation by $(3,-1)$ gives:
The new vertices are $(3,-1\pm7)$ and $(3\pm 4,-1)$ .
The new foci are $(3,-1\pm \sqrt{33})$.
