1. **Problem statement:** Find the volume of a cone with radius $r=4$ cm and height $h=8$ cm. 2. **Formula for volume of a cone:** $$V = \frac{1}{3} \pi r^2 h$$ This formula calculates the volume of a cone by taking one-third of the base area times the height. 3. **Calculate the volume for part (a):** Substitute $r=4$ and $h=8$ into the formula: $$V = \frac{1}{3} \pi (4)^2 (8)$$ $$V = \frac{1}{3} \pi \times 16 \times 8$$ $$V = \frac{1}{3} \pi \times 128$$ 4. **Simplify the fraction:** $$V = \cancel{\frac{1}{3}} \pi \times \cancel{128} \times \frac{128}{3} \pi$$ Actually, $128$ is not divisible by 3, so keep as: $$V = \frac{128}{3} \pi$$ 5. **Final volume for part (a):** $$V = \frac{128}{3} \pi \approx 134.04 \text{ cm}^3$$ 6. **Part (b) question:** If the height is doubled, new height $h' = 2 \times 8 = 16$ cm. 7. **Calculate new volume $V'$:** $$V' = \frac{1}{3} \pi r^2 h' = \frac{1}{3} \pi (4)^2 (16) = \frac{1}{3} \pi \times 16 \times 16 = \frac{256}{3} \pi$$ 8. **Compare new volume to original:** $$\frac{V'}{V} = \frac{\frac{256}{3} \pi}{\frac{128}{3} \pi} = \frac{256}{128} = 2$$ 9. **Conclusion:** Doubling the height doubles the volume of the cone. **Answer:** (a) Volume $= \frac{128}{3} \pi \approx 134.04$ cm$^3$ (b) Volume doubles when height is doubled.