Let's analyze the given diagram and the corresponding questions:

1. PS is a diameter of the circle, and as a property of circles, any angle subtended by the diameter at the circumference is a right angle (90°).
2. TRQ and TSP are straight lines, meaning angles on a straight line add up to 180°.
3. Given:
   • Angle $PQR = 108^\circ$
   • Angle $RTP = 24^\circ$

(a) We need to find:

1. Angle PRS

• Since $PS$ is the diameter, angle $PRS$ subtended by the diameter at $R$ will be a right angle: $\boxed{90^\circ} \quad \text{(Angle in a semicircle is } 90^\circ \text{)}$

2. Angle QPT

• Consider the quadrilateral $PQRT$ formed in the circle. The sum of angles in a cyclic quadrilateral is 180°.
• In quadrilateral $PQRT$, angles $PQR$ and $QPT$ are opposite. $\text{Thus, } PQR + QPT = 180^\circ$ $QPT = 180^\circ - 108^\circ = \boxed{72^\circ} \quad \text{(Opposite angles in a cyclic quadrilateral add up to } 180^\circ \text{)}$

3. Angle PSR

• This angle can be found by considering the fact that the angle $RTP = 24^\circ$, and angles on a straight line add up to 180°. $PSR = 180^\circ - 24^\circ = \boxed{156^\circ} \quad \text{(Angles on a straight line add up to } 180^\circ\text{)}$

4. Angle SRT

• Angle $SRT$ can be found similarly. Since $TRQ$ is a straight line and the angle $RTP = 24^\circ$, the remaining angle is: $SRT = 180^\circ - 156^\circ = \boxed{24^\circ} \quad \text{(Angles on a straight line add up to } 180^\circ\text{)}$

(b) Find angle QPR.

• Since angle $PQR$ is 108° and $PS$ is the diameter, $QPR$ is the remaining angle in triangle $PQR$. The sum of angles in a triangle is 180°: $QPR = 180^\circ - (PQR + PRS) = 180^\circ - (108^\circ + 90^\circ) = \boxed{12^\circ}$

Would you like further clarifications or a deeper explanation of any of these steps? Here are five related questions for further practice:

1. What is the sum of angles in any triangle?
2. How would you prove that angle PRS is 90° using circle theorems?
3. If $PQR = 90^\circ$, what would the angle $QPT$ be?
4. How would the answers change if $PQR$ was less than 108°?
5. What is the relationship between angles subtended by the same arc in a circle?

Tip: In cyclic quadrilaterals, opposite angles always sum to 180°. This property can help solve many angle problems in circle geometry.