(i) The volume of the tank is given by \(x \times \frac{1}{2}x \times h = 4\). Thus, \(\frac{1}{2}x^2h = 4\), giving \(h = \frac{8}{x^2}\).

The external surface area \(A\) is the sum of the areas of the sides and the lid:

\(A = x \times \frac{1}{2}x + 2(x \times h) + 2\left(\frac{1}{2}x \times h\right) + \frac{5}{4}x \times \frac{4}{5}x\).

\(A = \frac{1}{2}x^2 + 2xh + xh + \frac{5}{4}x \times \frac{4}{5}x\).

\(A = \frac{1}{2}x^2 + 3xh + \frac{1}{2}x^2\).

Substituting \(h = \frac{8}{x^2}\), we get:

\(A = \frac{3}{2}x^2 + 3x \times \frac{8}{x^2}\).

\(A = \frac{3}{2}x^2 + \frac{24}{x}\).

(ii) To find the minimum, differentiate \(A\) with respect to \(x\):

\(\frac{dA}{dx} = 3x - \frac{24}{x^2}\).

Set \(\frac{dA}{dx} = 0\):

\(3x - \frac{24}{x^2} = 0\).

\(3x^3 = 24\).

\(x^3 = 8\).

\(x = 2\).

To confirm it's a minimum, check the second derivative:

\(\frac{d^2A}{dx^2} = 3 + \frac{48}{x^3}\).

\(\frac{d^2A}{dx^2} > 0\) when \(x = 2\), hence \(A\) is a minimum.