(i) Differentiate \(y = (3x + 4)^{\frac{1}{2}}\) to find the gradient: \(\frac{1}{2}(3x + 4)^{-\frac{1}{2}} \times 3 = \frac{3}{2}(3x + 4)^{-\frac{1}{2}}\).

At \(x = 4\), \(\frac{dy}{dx} = \frac{3}{8}\).

The equation of the tangent is \(y - 3 = \frac{3}{8}(x - 4)\), which simplifies to \(y = \frac{3}{8}x + \frac{5}{2}\).

(ii) The area under the line is \(\frac{1}{2} \times 4 \times 3 = 3\).

The area under the curve is \(\int_0^4 (3x + 4)^{\frac{1}{2}} \, dx = \left[ \frac{2}{3}(3x + 4)^{\frac{3}{2}} \right]_0^4 = 13\).

The area of the shaded region is \(13 - 12 = \frac{5}{9}\) (or 0.556).

(iii) Given \(\frac{dy}{dx} = \frac{1}{2}\), solve \(\frac{3}{2}(3x + 4)^{-\frac{1}{2}} = 1\).

This gives \((3x + 4)^{\frac{1}{2}} = 3\), so \(3x + 4 = 9\), leading to \(x = \frac{5}{3}\).