The diagram shows a sector of a circle with centre O, radius 8 cm and sector angle 165 degrees. (a) Calculate the total perimeter of the sector. (b) The surface area of a sphere is the same as the area of the sector. Calculate the radius of the sphere.

The formula for arc length $L$ is $L = \frac{\theta}{360^\circ} \times 2\pi r$.
Given $\theta = 165^\circ$ and $r = 8$ cm.

$$L = \frac{165}{360} \times 2 \times \pi \times 8$$ $$L = \frac{165}{360} \times 16\pi$$ $$L = \frac{11}{24} \times 16\pi$$ $$L = \frac{11 \times 2}{3} \times \pi$$ $$L = \frac{22}{3}\pi \text{ cm}$$ $$L \approx 23.038 \text{ cm}$$

The perimeter of the sector is the sum of the arc length and two radii.
Perimeter $P = L + 2r$

$$P = \frac{22}{3}\pi + 2(8)$$ $$P = \frac{22}{3}\pi + 16$$ $$P \approx 23.038 + 16$$ $$P \approx 39.038 \text{ cm}$$

Rounding to one decimal place:

$$P \approx 39.0 \text{ cm}$$

The total perimeter of the sector is $\boxed{39.0 \text{ cm}}$.

The formula for the area of a sector $A_{sector}$ is $A_{sector} = \frac{\theta}{360^\circ} \times \pi r^2$.
Given $\theta = 165^\circ$ and $r = 8$ cm.

$$A_{sector} = \frac{165}{360} \times \pi \times 8^2$$ $$A_{sector} = \frac{165}{360} \times 64\pi$$ $$A_{sector} = \frac{11}{24} \times 64\pi$$ $$A_{sector} = \frac{11 \times 8}{3} \times \pi$$ $$A_{sector} = \frac{88}{3}\pi \text{ cm}^2$$ $$A_{sector} \approx 92.153 \text{ cm}^2$$

Let $r_s$ be the radius of the sphere. The surface area of the sphere is $A_{sphere} = 4\pi r_s^2$.

$$A_{sphere} = A_{sector}$$ $$4\pi r_s^2 = \frac{88}{3}\pi$$

Divide both sides by $\pi$:

$$4r_s^2 = \frac{88}{3}$$

Divide both sides by 4:

$$r_s^2 = \frac{88}{3 \times 4}$$ $$r_s^2 = \frac{88}{12}$$ $$r_s^2 = \frac{22}{3}$$ $$r_s = \sqrt{\frac{22}{3}}$$ $$r_s \approx \sqrt{7.33333}$$ $$r_s \approx 2.7080 \text{ cm}$$

Rounding to two decimal places (or 3 significant figures):

$$r_s \approx 2.71 \text{ cm}$$

The radius of the sphere is $\boxed{2.71 \text{ cm}}$.