Because there's a subtlety here, this question is worth a correct answer. But let's develop it with as little work as possible, in the most straightforward manner.
What subtlety? The variables $(U,V)$ do not determine $(X,Y).$
The change of variables from $(X,Y)$ to $(U,V)$ is two-to-one: because $(U,V)$ gives us information about $Y$ only in the form of $Y^2,$ whenever $(X,Y)$ corresponds to $(U,V),$ so does $(X,-Y).$ Almost surely, $Y\ne -Y$ (the chance of this for a Normal distribution of $Y$ is zero). This means the density of $(U,V)$ will be twice what a mindless application of the routine Calculus formulas indicates.
Those routine formulas are, of course, the computation of the Jacobian. This is just an old-fashioned term for computing the differential element $\mathrm{d}x\,\mathrm{d}y$ in terms of the new variables. One of the easiest ways to work it out starts with the formulas
$$X = V\sqrt{U};\ Y = \sqrt{1-V^2}\sqrt{U}.$$
Taking differentials,
$$\begin{aligned}
\mathrm{d}x\,\mathrm{d}y &= \left(\frac{v}{2\sqrt{u}}\mathrm{d}u + \sqrt{u}\mathrm{d}v\right)\,\left(\frac{\sqrt{1-v^2}}{2\sqrt{u}}\mathrm{d}u - \frac{v\sqrt{u}}{\sqrt{1-v^2}}\mathrm{d}v\right) \\
&= \frac{1}{2\sqrt{1-v^2}}\mathrm{d}v\,\mathrm{d}u.
\end{aligned}$$
Substitute everything into the original probability element for the bivariate Normal distribution taking care to indicate what the possible values of the variables $u$ and $v$ can be. Omitting this is another pitfall that plagues the uninitiated, so I will be explicit, using $\mathcal{I}$ to represent the indicator function:
$$\begin{aligned}
f_{X,Y}(x,y)\,\mathrm{d}x\,\mathrm{d}y &= \frac{1}{2\pi\sigma^2} \exp\left(-\frac{x^2+y^2}{2\sigma^2}\right)\,\mathrm{d}x\,\mathrm{d}y \\
&= \frac{1}{2\pi\sigma^2} \exp\left(-\frac{u}{2\sigma^2}\right)\, \frac{1}{2\sqrt{1-v^2}}\mathrm{d}v\,\mathrm{d}u\ \mathcal{I}(u\ge 0)\,\mathcal{I}(-1\le v\le 1).
\end{aligned}$$
(It would be incorrect to write this formula without the indicator functions. Usually readers are expected to notice that $\sqrt{1-v^2}$ is defined only for $-1\le v\le 1,$ so in informal settings we can get away without indicating that explicitly; but it's not quite as noticeable that although $\exp(-u/(2\sigma^2))$ is defined everywhere, its integral diverges unless $u$ is explicitly restricted.)
Introduce the factor of $2$ from the two-to-one transformation and notice the probability element splits into a factor depending only on $u$ and one depending only on $v:$
$$f_{U,V}(u,v)\,\mathrm{d}v\,\mathrm{d}u = \left[\frac{1}{2\sigma^2} \exp\left(-\frac{u}{2\sigma^2}\right)\,\mathrm{d}u\ \mathcal{I}(u\ge 0)\right]\, \left[\frac{1}{\pi\sqrt{1-v^2}}\,\mathrm{d}v\ \mathcal{I}(-1\le v\le 1)\right].$$
In one stroke this answers (a) and (b): because the probability element factors, the variables $U$ and $V$ are independent.
As a check, you may integrate the two factors separately over the set of real numbers: each integrates to $1,$ as it must for any univariate probability distribution.
Question (c) is a routine exercise in a univariate change of variable, so discussing it doesn't add any further interest.
