Consider the transformation $T:[0,1]^2\to \mathbb R^2$ defined by $$ T(x,y)=(\cos(2\pi y)\sqrt{-2\log x},\sin(2\pi y)\sqrt{-2\log x}). $$ Then, for every $(u,v)$ in $\mathbb R^2\setminus\{(0,0)\}$, $$ f_{U,V}(u,v)=f_{X,Y}(T^{-1}(u,v))\cdot|\det J_T(T^{-1}(u,v))|^{-1}, $$ provided $T$ is bijective and the Jacobian matrix $J_T$ is nonsingular on $[0,1]^2$. This remark allows to skip the computation of $T^{-1}$. Your next step is to compute the $2\times2$ matrix $J_T$, to deduce its determinant, and to plug this value into the expression of $f_{U,V}$.
A shortcut the physicists (and others) use is to rely on the formalism of differential forms. That is, one starts from $$ u=\cos(2\pi y)\sqrt{-2\log x},\quad v=\sin(2\pi y)\sqrt{-2\log x}, $$ and one differentiates $u$ and $v$. This yields $$ \mathrm du=-2\pi\sin(2\pi y)\sqrt{-2\log x}\mathrm dy-\frac{\cos(2\pi y)}{x\sqrt{-2\log x}}\mathrm dx, $$ and $$ \mathrm dv=+2\pi\cos(2\pi y)\sqrt{-2\log x}\mathrm dy-\frac{\sin(2\pi y)}{x\sqrt{-2\log x}}\mathrm dx. $$ Now, $\mathrm du\mathrm dv$ is the product of these, with the convention that $\mathrm du\mathrm du=\mathrm dv\mathrm dv=0$ and that $\mathrm dv\mathrm du=-\mathrm du\mathrm dv$. In other words, if $\mathrm du=\alpha\mathrm dx+\beta\mathrm dy$ and $\mathrm du=\gamma\mathrm dx+\delta\mathrm dy$, then $$ \mathrm du\mathrm dv=|\alpha\delta-\beta\gamma|\,\mathrm dx\mathrm dy. $$ Using either the mathematician's way or the physicist's way, in the present case one gets $$ \mathrm du\mathrm dv=\frac{2\pi}x\,\mathrm dx\mathrm dy, $$ hence $$ |\det J_T(x,y)|^{-1}=\frac{x}{2\pi}. $$ A last remark is that, if $T(x,y)=(u,v)$, then $x=\exp(-(u^2+v^2)/2)$, hence $$ f_{U,V}(u,v)=\frac{\mathrm e^{-(u^2+v^2)/2}}{2\pi}. $$ From here, the other questions you were asked should be direct.