= Solution
Define $T^\dagger(0)=0$; the printed quotient is undefined at the origin, but this continuous extension changes no transport cost because $\mu(\{0\})=0$. In <polar coordinates>, the two <probability density functions> give
$$
\mu\{|x|\leq r\}=r^2,\qquad
\nu\{|y|\leq s\}=s^4\quad(0\leq r,s\leq1).
$$
Both angular distributions are uniform, with <independence> of angle and radius. The proposed <transport map> preserves the angle and sends $r$ to $s=\sqrt r$. Therefore
$$
\mu\{|T^\dagger(x)|\leq s\}=\mu\{|x|\leq s^2\}=s^4,
$$
and preservation of the angle proves $(T^\dagger)_\#\mu=\nu$. As a local check using the <Jacobian determinant>, its radial and tangential derivatives for $r>0$ are $1/(2\sqrt r)$ and $1/\sqrt r$, respectively, so $\det DT^\dagger=1/(2r)$ and $g(T^\dagger(x))\det DT^\dagger(x)=(2r/\pi)/(2r)=f(x)$.
Now use the <convex function>
$$
\Phi(x)=\frac23|x|^{3/2}\quad(x\in\mathbb R^2).
$$
It is <convex> because the <Euclidean norm> is <convex> and $s\mapsto(2/3)s^{3/2}$ is increasing and <convex> on $[0,\infty)$. It is <differentiable>, including at zero, and
$$
\boxed{\nabla\Phi(x)=T^\dagger(x)=\frac{x}{\sqrt{|x|}}\ (x\ne0),\qquad\nabla\Phi(0)=0.}
$$
Its graph <transport plan> lies in the graph of $\partial\Phi$, so the <Knott–Smith optimality criterion> proves quadratic optimality. Since that plan is induced by a map, part 1(b) proves optimality for the <Monge optimal transport problem> as well. The minimum provides a useful independent check:
$$
\boxed{\min\mathbb M=\int_0^1(r-\sqrt r)^2\,2r\,dr
=\frac12-\frac87+\frac23=\frac1{42}.}
$$
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