(a) To find the rate at which \(h\) is increasing, we use the chain rule. First, differentiate \(V\) with respect to \(h\):

\(\frac{dV}{dh} = \frac{4}{3} \times 3(25 + h)^2 = 4(25 + h)^2\).

At \(h = 10\), \(\frac{dV}{dh} = 4(25 + 10)^2 = 4900\).

Using the chain rule, \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\).

Given \(\frac{dV}{dt} = 500\), we have:

\(500 = 4900 \cdot \frac{dh}{dt}\).

Solving for \(\frac{dh}{dt}\), we get \(\frac{dh}{dt} = \frac{500}{4900} = 0.102 \text{ cm/s}\).

(b) Given \(\frac{dh}{dt} = 0.075 \text{ cm/s}\), use the chain rule:

\(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} = 4(25 + h)^2 \cdot 0.075\).

Set \(\frac{dV}{dt} = 500\):

\(500 = 4(25 + h)^2 \cdot 0.075\).

Solving for \(h\), we have:

\((25 + h)^2 = \frac{5000}{3}\).

\(25 + h = \sqrt{\frac{5000}{3}}\).

\(h = \sqrt{\frac{5000}{3}} - 25 \approx 15.82\).

Substitute \(h\) back into the volume formula:

\(V = \frac{4}{3}(25 + 15.82)^3 - \frac{62500}{3}\).

\(V \approx 69900 \text{ cm}^3\).