a) $A, B, O, T$ are clearly concyclic since radii $OA$ and $OB$ are perpendicular to the tangents. As for $S$, note that $OS$ bisects $\angle BSC$, so $\angle OSB = \angle OSC = 90^{\circ} - \angle C$. Now we know $\angle BOC = 2\angle C$. Since $\bigtriangleup BOC$ is isoceles on $O$, we get that $\angle OBC = \angle 90^{\circ} - \angle C = \angle OSC$, implying $S$ is also concyclical with the other 4 points of interest. b) Concyclicity and $\angle TBO=90^{\circ}$ implies $\angle OST = 90^{\circ}$, i.e $ST\perp OS$. But $OS$ is an altitude within $\bigtriangleup BOC$, so $OS\perp BC$. This means $ST\parallel BC$ as desired.

Let $D$ be the intersection point of the line through $A$ parallel to $\overline{BC}$ and the circumcircle of $\triangle ABC$. It suffices to show that $\overline{ST}, \overline{AD}, \overline{BC}$ concur at a point at $\infty$. Take the projective transformation mapping $\triangle ABC$ to a right triangle $\triangle A'B'C'$ preserving the circumcircle with $A'C'$ being a diameter. Observe then that $A'B'C'D'$ is a rectangle, $S'=\overline{A'C'}\cap\overline{B'D'}$ is a point at $\infty$, and that $T'$ is also a point at $\infty$ since the tangents at $A$ and $B$ are parallel. Then $S'T'$ is the line at $\infty$, and therefore passes through $\overline{A'D'}\cap\overline{B'C'}$, as desired. $\blacksquare$