(a) To find angle BOC, we use the formula for arc length: \(s = r\theta\). Given the arc length AB is 2 cm and radius is 10 cm, we have:

\(\theta = \frac{s}{r} = \frac{2}{10} = 0.2 \text{ radians}\)

Angle BOC is the sum of angle AOC and angle AOB:

\(\angle BOC = \angle AOC + \angle AOB = \frac{1}{6} \pi + 0.2 \approx 0.724 \text{ radians}\)

(b) To find the area of the shaded region BPC, we first find the lengths of BP and OP using trigonometry in triangle OBP:

\(BP = 10 \sin \left( \frac{5\pi + 6}{30} \right) \approx 6.6208\)

\(OP = 10 \cos \left( \frac{5\pi + 6}{30} \right) \approx 7.494\)

The area of triangle OBP is:

\(\text{Area of } \triangle OBP = \frac{1}{2} \times 10 \times 10 \times \sin \left( \frac{5\pi + 6}{30} \right) \approx 24.809 \text{ cm}^2\)

The area of sector BOC is:

\(\text{Area of sector BOC} = \frac{1}{2} \times 10^2 \times \left( \frac{5\pi + 6}{30} \right) \approx 36.1799 \text{ cm}^2\)

The area of the shaded region BPC is:

\(\text{Area of region BPC} = \text{Area of sector BOC} - \text{Area of } \triangle OBP \approx 11.4 \text{ cm}^2\)