(i) To find \(\frac{dy}{dx}\), we differentiate each term of \(y = 4\sqrt{x} + \frac{2}{\sqrt{x}}\).

Rewrite \(y = 4x^{\frac{1}{2}} + 2x^{-\frac{1}{2}}\).

Differentiate: \(\frac{dy}{dx} = 4 \cdot \frac{1}{2}x^{-\frac{1}{2}} + 2 \cdot (-\frac{1}{2})x^{-\frac{3}{2}}\).

Simplify: \(\frac{dy}{dx} = 2x^{-\frac{1}{2}} - x^{-\frac{3}{2}}\).

(ii) Use the chain rule: \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).

Given \(\frac{dx}{dt} = 0.12\) and \(x = 4\), substitute into \(\frac{dy}{dx} = 2x^{-\frac{1}{2}} - x^{-\frac{3}{2}}\).

At \(x = 4\), \(\frac{dy}{dx} = 2(4)^{-\frac{1}{2}} - (4)^{-\frac{3}{2}} = \frac{1}{\sqrt{4}} - \frac{1}{8} = \frac{1}{2} - \frac{1}{8} = \frac{3}{8}\).

Thus, \(\frac{dy}{dt} = \frac{3}{8} \times 0.12 = 0.105\).