(i) To find the volume \(V\) of water in the tank, we first determine the area of the triangle \(ABE\). Since \(\angle ABE = \angle BAE = 30^\circ\), the triangle is isosceles with \(\angle AEB = 120^\circ\). Using the tangent of \(60^\circ\), we have:

\(\tan 60 = \frac{x}{h} \Rightarrow x = h \tan 60\)

\(\tan 60 = \sqrt{3}\), so \(x = h\sqrt{3}\).

The area \(A\) of triangle \(ABE\) is:

\(A = \frac{1}{2} \times x \times h = \frac{1}{2} \times h\sqrt{3} \times h = \frac{h^2\sqrt{3}}{2}\)

The volume \(V\) of the prism is:

\(V = \text{Area of } ABE \times AD = \frac{h^2\sqrt{3}}{2} \times 40 = 20h^2\sqrt{3}\)

Thus, \(V = (40\sqrt{3})h^2\).

(ii) Given \(\frac{dV}{dt} = 200\) cm\(^3\) s\(^{-1}\), we find \(\frac{dh}{dt}\) when \(h = 5\).

Differentiate \(V = (40\sqrt{3})h^2\) with respect to \(h\):

\(\frac{dV}{dh} = 80h\sqrt{3}\)

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

\(200 = 80h\sqrt{3} \cdot \frac{dh}{dt}\)

When \(h = 5\):

\(200 = 80 \times 5 \times \sqrt{3} \cdot \frac{dh}{dt}\)

\(\frac{dh}{dt} = \frac{200}{400\sqrt{3}} = \frac{1}{2\sqrt{3}}\)

Thus, \(\frac{dh}{dt} = \frac{1}{2\sqrt{3}}\) or approximately 0.289.