Answer: a) $90^{\circ}$, b) $2: 1$, c) $V=15$.

Solution. a) Points $P$ and $Q$ lie on the circle with diameter $BC$; therefore, $\angle BPC=90^{\circ}, \angle BQC=90^{\circ}$ (i.e., $BP$ and $CQ$ are altitudes of triangle $ABC$). The line $BP$ is the projection of the line $TP$ onto the base plane, and since $BP \perp PA$, by the theorem of three perpendiculars, $TP \perp PA$, i.e., $\angle TPA=90^{\circ}$.

b) Since the lines $B_1P$ and $C_1Q$ intersect, all four points $B_1, C_1, P$, and $Q$ lie in the same plane (let's call it $\alpha$). Therefore, the lines $PQ$ and $B_1C_1$ lie in the same plane $\alpha$, and since they do not intersect (as they lie in the parallel bases of the prism), $PQ \parallel B_1C_1$. Hence, $PQ \parallel BC$. The trapezoid $PQBC$ is inscribed in a circle, so it is isosceles, and the angles at its base $BC$ are equal, which means triangle $ABC$ is isosceles ($AB=AC$).

Triangles $B_1C_1T$ and $PQT$ are similar by two angles. From the equality of triangles $CC_1Q$ and $BB_1P$, it follows that $B_1P = C_1Q$, so both triangles $B_1C_1T$ and $PQT$ are isosceles with bases $B_1C_1$ and $PQ$ respectively. Therefore, $PQ: BC = PQ: B_1C_1 = TQ: TC_1 = TQ: (QC_1 - TQ) = TQ: (B_1P - TQ) = 2: 3$, from which $\frac{CP}{AP} = \frac{AC - AP}{AP} = \frac{AC}{AP} - 1 = \frac{BC}{PQ} - 1 = \frac{3}{2} - 1 = \frac{1}{2}$.

c) If $AC = 3$, then $CP = 1, AP = 2, AB = 3; BP = \sqrt{AB^2 - AP^2} = \sqrt{5}; BB_1 = \sqrt{B_1P^2 - BP^2} = 2\sqrt{5}$. Therefore, the area of the base of the prism is $S_{ABC} = \frac{1}{2} \cdot 3 \cdot \sqrt{5}$, and the volume of the prism is $V = S_{ABC} \cdot BB_1 = 15$.
7. Given a right triangular prism $A B C A_{1} B_{1} C_{1}$. A sphere with diameter $B C$ intersects the edges $A C$ and $A B$ at points $P$ and $Q$, respectively, different from the vertices of the prism. Segments $B_{1} P$ and $C_{1} Q$ intersect at point $T$, and $B_{1} P=5, T Q=2$.

a) Find the angle $T P A$.

b) Find the ratio $A P: C P$.

c) Suppose it is additionally known that $A C=3$. Find the volume of the prism.