(a) The rate of change of volume is given by \(\frac{dV}{dt} = -k\sqrt{h}\). The volume of the hemisphere is \(V = \frac{1}{3} \pi (3rh^2 - h^3)\). Differentiating with respect to h, \(\frac{dV}{dh} = 2\pi rh - \pi h^2\).

Using the chain rule, \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\), we have \(-k\sqrt{h} = (2\pi rh - \pi h^2) \cdot \frac{dh}{dt}\).

Rearranging gives \(\frac{dh}{dt} = -\frac{k\sqrt{h}}{2\pi rh - \pi h^2}\). Let \(B = k\pi\), then \(\frac{dh}{dt} = -\frac{B}{2rh^2 - h^3}\).

(b) Separate variables: \((2rh^2 - h^3) dh = -B dt\). Integrate both sides: \(\int (2rh^2 - h^3) dh = -\int B dt\).

The left side integrates to \(\frac{2}{3}rh^3 - \frac{1}{4}h^4\) and the right side to \(-Bt + C\).

At \(t = 0, h = r\), so \(\frac{2}{3}r^4 - \frac{1}{4}r^4 = C\).

At \(t = 14, h = 0\), so \(0 = -14B + C\).

Solving these gives \(C = \frac{5}{12}r^4\) and \(B = \frac{1}{15}r^2\).

Substitute back to find \(t\): \(t = 14 - 20\left(\frac{h}{r}\right)^{\frac{3}{2}} + 6\left(\frac{h}{r}\right)^{\frac{5}{2}}\).