The line $y=a^2 x$ and the curve $y=x(b-x)^2$, where $0<a<b$, intersect at the origin $O$ and at points $P$ and $Q $. The $x$-coordinate of $P$ is less than the $x$-coordinate of $Q$.

1. Find the coordinates of $P$ and $Q$, and sketch the line and the curve on the same axes.
2. Show that the equation of the tangent to the curve at $P$ is $$ y = a(3a-2b)x + 2a(b-a)^2. $$
3. This tangent meets the $y$-axis at $R$. The area of the region between the curve and the line segment $OP$ is denoted by $S$. Show that $$ S= \frac{1}{12}(b-a)^3(3a+b). $$
4. The area of triangle $OPR$ is denoted by $T$. Show that $S>\frac{1}{3}T$.