Imagine the complex numbers identified with the plane $\mathbb{R}^2$, by letting a complex number $a+bi$ correspond to the point $(a,b)$.
If we now translate to polar coordinates, so that $(a,b)$ corresponds to $(r,\theta)$ (where $a=r\cos\theta$ and $b=r\sin\theta$), then we can also express the number $a+bi$ as $r(\cos\theta + i\sin\theta).$ This, in turn, can be written as $re^{i\theta},$ since $e^{i\theta}=\cos\theta + i\sin\theta$ when $\theta$ is a real number.
De Moivre's Theorem says that if we write $a+bi = r(\cos\theta + i\sin\theta)$ as above, then $(a+bi)^n = r^n\Bigl(\cos(n\theta) + i\sin(n\theta)\Bigr).$ (An easy of seeing this is to use the complex exponential, since $(a+bi)^n = (re^{i\theta})^n = r^n(e^{i\theta})^n = r^n e^{in\theta},$ but this would require you to prove that the complex exponential has the same property as the real exponential that $(e^u)^v = e^{uv}$, so it might be a bit circular.)
More generally, if $a+bi = r\bigl(\cos\theta + i\sin\theta\bigr)$ and $c+di = s\bigl(\cos\phi + i\sin\phi)$, then $(a+bi)(c+di) = rs\bigl(\cos(\theta+\phi) + i\sin(\theta+\phi)\bigr).$ That is: you can multiply complex numbers by multiplying their norm (size) and adding their argument.
To solve the first problem, convert $3+3i$ into polar form, and use De Moivre's Theorem; then convert back.
To solve the second problem, convert $1+i\sqrt{3}$ into polar form, and figure out what numbers you need for $r$ and $\theta$ so that $r^2(\cos(2\theta) + i\sin(2\theta))$ equals that result. Similarly for $3$.
For the fourth problem, use the quadratic formula (and use a method like that of problem 2 to figure out the complex values of the square roots you get).