A and B were fine, for C I took componentwise vectors x(t) and y(t), differentiated twice and applied $1/2mv^2$.
For (d) I have no idea, and for (e) I applied:
$1/2m(v_2^2-v_1^2)$;
From then on, I have very little idea of what to do.
I would be very grateful of any help.
If you have done B, you get an expression for force which is directed towards the Origin ($F=m\vec{\ddot{r}}=-m\omega^2\vec{r}$). Take the dot product of this with $d\vec{r}$. \begin{equation} dW=-m\omega^2\vec{r}.\vec{dr}=-m\omega^2(xdx+ydy) \end{equation} Integrating over the line from A to B gives the work asked in (d). \begin{equation} W=-m\omega^2\frac{(x^2+y^2)|^{B}_{A}}{2}=\frac{1}{2}m\omega^2(r_A^2-r_B^2) \end{equation} where $r=|\vec{r}|$.
This, along with the difference in kinetic energy which you have calculated proves (e)
(e) shows that work done in moving the particle from one point to another is dependent only on the kinetic energy at the end points, and thus irrespective of the path traversed. This proves (f) and (g).