(i) Differentiate \(y = 2x + \frac{8}{x^2}\) with respect to \(x\):

\(\frac{dy}{dx} = 2 - \frac{16}{x^3}\).

Differentiate again to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{48}{x^4}\).

(ii) Set \(\frac{dy}{dx} = 0\) to find the stationary point:

\(2 - \frac{16}{x^3} = 0 \Rightarrow x = 2\).

Substitute \(x = 2\) into the original equation:

\(y = 2(2) + \frac{8}{2^2} = 4 + 2 = 6\).

Stationary point is \((2, 6)\).

Since \(\frac{d^2y}{dx^2}\) is positive, the stationary point is a minimum.

(iii) At \(x = -2\), \(\frac{dy}{dx} = 4\), so the gradient of the tangent is 4.

The gradient of the normal is \(-\frac{1}{4}\).

Equation of the normal: \(y + 2 = -\frac{1}{4}(x + 2)\).

Set \(y = 0\) to find the x-intercept:

\(0 + 2 = -\frac{1}{4}(x + 2) \Rightarrow x = -10\).

The normal intersects the x-axis at \((-10, 0)\).