(i) At \(x = 4\), \(\frac{dy}{dx} = 2\).

Using the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = 2 \times 3 = 6\).

(ii) Integrate \(\frac{dy}{dx} = 1 + 2x^{-\frac{1}{2}}\) to find \(y\).

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

Substitute \(x = 4, y = 6\):

\(6 = 4 + 4(4^{\frac{1}{2}}) + c\)

\(c = -6\)

Thus, \(y = x + 4x^{\frac{1}{2}} - 6\).

(iii) Equation of the tangent at A is \(y - 6 = 2(x - 4)\) or \(y = 2x - 8\).

The tangent crosses the x-axis at \(B = (1, 0)\).

The gradient of the normal is \(-\frac{1}{2}\), so the equation is \(y = -\frac{1}{2}x + 8\).

The normal crosses the x-axis at \(C = (16, 0)\).

The area of triangle ABC is \(\frac{1}{2} \times 15 \times 6 = 45\).