Answer
$(3\displaystyle \sqrt{2}, \frac{3\pi}{4}),\ (-3\displaystyle \sqrt{2}, \frac{7\pi}{4})$
Work Step by Step
The point P, with rectangular coordinates ($-3,3)$ lies in the 3rd quadrant.
Change from rectangular to polar coordinates with
$r^{2}=x^{2}+y^{2}$ and $\displaystyle \tan\theta=\frac{y}{x}$.
$r^{2}=3^{2}+3^{2}=18$
$r=\pm 3\sqrt{2}$
$\displaystyle \tan\theta=\frac{3}{-3}=-1,$ and
with $\theta$ in quadrant II, $\displaystyle \theta=\frac{3\pi}{4}$, r is positive:
so the appropriate polar coordinates are $(3\displaystyle \sqrt{2}, \frac{3\pi}{4}).$
With $\theta$ in quadrant IV, $\displaystyle \theta=\frac{7\pi}{4}$, r is negative:
so the appropriate polar coordinates are $(-3\displaystyle \sqrt{2}, \frac{7\pi}{4}).$