This is for part a. Suppose that a polarizing $f$ is surjective. There is at least one point $P$ such that $P \not \in f(P)$. To see this, suppose the opposite: that is, suppose $P \in f(P)$ for all point $P$. Let $P$ be any point. Then for every $Q \in f(P)$ where $Q \neq P$, it follows that $P \in f(Q)$. However $f(Q)$ passes through $Q$. Hence $f(P) = f(Q)$. Now consider a line through $P$ that does not pass through $Q$. By surjectivity the line is $f(R)$ for some $R \neq P$. But this implies, by the same argument, $f(R) = f(P)$ which is an absurdity. Therefore there is at least one point $P$ such that $P \not \in f(P)$. Now, with this point $P$, $P \not \in f(P)$, consider $f(P)$ and a line parallel to $f(P)$ and passing through $P$, say, by surjectivity again, $f(Q)$. Since $P \in f(Q)$, we have $Q \in f(P)$. Now consider the line $PQ$, say $f(S)$. Then $S \in f(P)$ at the same time $S \in f(Q)$ which is impossible. Since all the possibilities are elliminated, we have that there is no surjective polarizing function.

This is for part b. An example is the pole-polar relation respect to a parabola.

An interesting version of this is to think what happens when $$f:\mathbb{R}^2-\{(0,0)\}\rightarrow \mathcal{H}*$$where $\mathcal{H}*$ denotes the lines in the plane that does not go through the origin. The solution above no longer works so we may have surjective functions now. In fact, an example is pole-polar wrt a circle whose center is the origin. The question is, are there any other such functions?