Solving the problem hinges on finding the answer to part (a). A geometric approach will be described.
There are 3 cases. If $0 \le r \le 1$, the probability is just $1/4$ the area of a circle of radius $r$, so it is $\pi r^2/4$.
If $1 things are messy. Draw a picture! We have a circle, part of which lies inside the square. We want to find the area of the part of the circle that is inside the square, and then divide by $4$.
The area of the part of the circle inside the square is the area of the circle, minus the sum of the areas of the $4$ parts that stick out of the square.
Look at one of those parts. It is a sector of a circle, with angle $2\arccos(1/r)$, minus a triangle with base $2\sqrt{r^2-1}$ and height $1$.
So the area of the $4$ parts that stick out is $4\left[r^2\arccos(1/r)-\sqrt{r^2-1}\right]$ Subtract from $\pi r^2$, divide by $4$. We get $\pi r^2/4 + \sqrt{r^2-1} -r^2\arccos(1/r)$ Finally, the easiest case: if $r>\sqrt{2}$, our probability is $1$.
To find the cdf of the random variable $Z=X^2+Y^2$, we want to find an expression for the probability that $Z \le z$. We get this by setting $r=\sqrt{z}$ in the answers to part (a).
After all this work, the pdf is (sort of) easy. Differentiate the cdf of $Z$.