1. Yes
2. We can view the function $f$ as a continuous function on the complex plane defined by $f(\theta,t)=te^{i\theta}$. From this viewpoint, we can see that the image of this function is an open annulus minus the positive real line $$A=\{z\in\mathbb{C} : 0<|z|<1\}-\{x\in\mathbb{R} : 0<x<1 \}$$
($t$ acts as the radius which is in the interval $(0,1)$ and $\theta$ acts as the angle which vary in the interval $(0,2\pi)$. Pictorially, it is the pac-man shaped set (when he shut his mouth).
Thus the closure of the image of $f$ is the unit disk $D^2=\{z\in\mathbb{C} : |z|\leq 1\}$. To find the closure, you can just add the boundary of this set.
