(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{16}{x^3}\):

\(y = \int \frac{16}{x^3} \, dx = \int 16x^{-3} \, dx = \frac{16x^{-2}}{-2} + c = \frac{-8}{x^2} + c\).

Using the point \((1, 4)\), substitute to find \(c\):

\(4 = \frac{-8}{1^2} + c \Rightarrow c = 12\).

Thus, the equation of the curve is \(y = \frac{-8}{x^2} + 12\).

(ii) The normal has gradient \(-\frac{1}{2}\), so the perpendicular gradient is 2. Set \(\frac{16}{x^3} = 2\) to find \(x\):

\(16 = 2x^3 \Rightarrow x = 2\).

Substitute \(x = 2\) into the curve equation to find \(y\):

\(y = \frac{-8}{2^2} + 12 = 10\).

The equation of the normal is \(y - 10 = -\frac{1}{2}(x - 2)\).

Simplify to \(2y + x = 22\).

(iii) Find the area under the curve from \(x = 1\) to \(x = 2\):

\(A = \int_{1}^{2} \left( \frac{-8}{x^2} + 12 \right) \, dx\).

\(= \left[ \frac{8}{x} + 12x \right]_{1}^{2}\).

\(= \left( \frac{8}{2} + 12 \times 2 \right) - \left( \frac{8}{1} + 12 \times 1 \right)\).

\(= (4 + 24) - (8 + 12) = 28 - 20 = 8\).