(a)
(i) Using volume of reservoir = volume of hemisphere + volume of cone.
\(\implies 333\frac{1}{3}\pi = \frac{2}{3}\pi r^{3} + \frac{1}{3}\pi r^{2} h\)
\(\implies \frac{1000}{3}\pi = \frac{1}{3}\pi (2r^{3} + r^{2}h)\)
Let the radius of the hemisphere be x m;
Then \(\frac{1000}{3} = \frac{1}{3}(2x^{3} + x^{2}(6x))\)
\(1000 = 2x^{3} + 6x^{3}\)
\(1000 = 8x^{3} \implies x^{3} = \frac{1000}{8} = 125\)
\(x = \sqrt[3]{125} = 5 m\)
Hence, the volume of the hemisphere = \(\frac{2}{3}\pi r^{3}\)
(where r = x = 5 m)
= \(\frac{2}{3} \times \frac{22}{7} \times 5^{3} m^{3}\)
= \(\frac{5500}{21} m^{3} \approxeq 261.905 m^{3}\)
= \(262 m^{3} \) (nearest whole number)
(ii) Total surface area of reservoir
= surface area of hemisphere + surface area of cone
= \(2\pi r^{2} + \pi rl\)

From the diagram above, \(l^{2} = 30^{2} + 5^{2}\)
\(l^{2} = 900 + 25 = 925\)
\(l = \sqrt{925} = 30.41 m\)
Hence, the total surface area of the reservoir = \(2 \times \frac{22}{7} \times 5 \times 5 + \frac{22}{7} \times 5 \times 30.41\)
= \(157.143 + 477.871\)
= \(635.104 m^{2}\)
\(\approxeq 635 m^{2}\) (to the nearest whole number).