$\mathbf{r}(t)=((2+\cos 3t)\cos 2t, (2+\cos 3t)(\sin 2t)), t\in [0,2\pi ]$
Cartesian coordinates are getting me nowhere here, but i've no idea where to start with polar coordinates on this one since there is cos(3t) and cos(2t) so what would the value of theta be in order to eliminate t?
There are really two questions in this post, I'm going to address the self-intersection question and leave the drawing to someone else.
Suppose that $\vec{r}(t_1)=\vec{r}(t_2)$. Then, you know that $$ \|\vec{r}(t_1)\|=\|\vec{r}(t_2)\|. $$ We can calculate \begin{align*} \|\vec{r}(t)\|&=\sqrt{(2+\cos(3t))^2\cos^2(2t)+(2+\cos(3t))^2\sin^2(2t)}\\ &=\sqrt{(2+\cos(3t))^2(\cos^2(2t)+\sin^2(2t))}\\ &=\sqrt{(2+\cos(3t))^2}\\ &=|2+\cos(3t)|. \end{align*} Since the smallest that $\cos(3t)$ can be is $-1$, we can drop the absolute value bars and get that the length of the vector is $2+\cos(3t)$.
Therefore, the equality from above implies that $$ 2+\cos(3t_1)=2+\cos(3t_2) $$ or that $$ \cos(3t_1)=\cos(3t_2). $$ Then, you have two possibilities, either $$ 3t_1=3t_2+2k\pi $$ for some integer $k$, or $$ 3t_1=-3t_2+2k\pi $$ for some integer $k$.
Now, we have concluded that one of the following holds: \begin{align*} t_1&=t_2+\frac{2}{3}k\pi\\ t_1&=-t_2+\frac{2}{3}k\pi. \end{align*}
Let's look at the two cases one at a time, starting with the first one. In this case, $t_1=t_2+\frac{2}{3}k\pi$. By substituting these into the formula for $\vec{r}(t)$, we already found that $2+\cos(3t_1)=2+\cos(3t_2)$ (and these are nonzero), so what we really need is \begin{align*} \cos(2t_1)&=\cos(2t_2)\\ \sin(2t_1)&=\sin(2t_2). \end{align*} For the first equality to hold, you need $$ 2t_1=2t_2+2\ell\pi $$ or $$ 2t_1=-2t_2+2\ell\pi. $$ Substituting our formula for $t_1$ into the first equation gives $$ 2\left(t_2+\frac{2}{3}k\pi\right)=2t_2+2\ell\pi. $$ This simplifies to $$ \frac{4}{3}k\pi=2\ell\pi $$ or that $$ 2k=3\ell. $$ In the integers, the smallest positive value $k$ could be is $3$, but then $t_1$ and $t_2$ differ by $2\pi$, and hence must be $t_1=2\pi$ and $t_2=0$ (which is not very exciting).
On the other hand, if the second equality holds, by substituting in our formula for $t_1$, we get $$ 2\left(t_2+\frac{2}{3}k\pi\right)=-2t_2+2\ell\pi. $$ This simplifies to $$ t_2=\frac{1}{2}\left(\ell-\frac{2}{3}k\right)\pi. $$ Since $\ell$ and $k$ must be integers (and they can't be too large because $t_1$ and $t_2$ differ by at most $2\pi$), there are only a few possible values for $t_2$, which can be checked in the second equality (with sines) one-by-one.
Working through the second possibility for the relationship between $t_1$ and $t_2$ may give additional solutions, but I'll leave the details to you.