Solution:

Let $T$ be the circle circumscribed about the dodecagon, and let $O$ be its center; let $O_{S}$ be the center of the circle $S$ and let $Q$ be the midpoint of $A H$. The points $O$ and $O_{S}$ lie on the perpendicular bisector of $A H$ and, having the same distance from $A$ and $H$, are symmetric with respect to $A H$, and also symmetric with respect to $Q$. Since $A$ and $G$ are opposite vertices of the dodecagon, $A G$ is a diameter of $T$; it follows that $O$ is also the midpoint of the diagonal $A G$ and that $A \widehat{H} G=90^{\circ}$. By the similarity of triangles $A Q O$ and $A H G$ (two right triangles sharing an acute angle), we have $Q O=\frac{1}{2} H G$ and hence $O_{S} O=2 Q O=H G$.

The circle $S$ is thus the translate of $T$ by a length equal to the side $H G$, in the direction $\overrightarrow{G H}$.

(b): By the regularity of the dodecagon, $H L=N L$ and $G L=A L$ and therefore the lines $H N$ and $A G$, both perpendicular to the radius $L O$ of $T$, are parallel. But then $H N$ is the translate of $A G$ with respect to the vector $\overrightarrow{G H}$, and therefore the center of $S$ lies on $H N$.

(a) and (c): By what was observed before, if we show that $P$ is the translate of $E$ in the direction $\overrightarrow{G H}$, we simultaneously prove (a) and (c). Equivalently, we show that $E$ is the translate of $P$ in the direction $\overrightarrow{H G}$. The translate of $P$ in the direction $\overrightarrow{H G}$ is the intersection of the translates, in the same direction, of $A F$ and $B D$. But with the same argument used to prove (b) one shows that the translates of $A F$ and $D H$ in the direction $\overrightarrow{H G}$ are the diagonals $B E$ and $G E$, which intersect at $E$.