(i) Differentiate \(y = 5 - \frac{8}{x}\) to find the gradient of the tangent: \(\frac{dy}{dx} = \frac{8}{x^2}\).

At \(x = 2\), \(\frac{dy}{dx} = \frac{8}{4} = 2\). The gradient of the normal is \(-\frac{1}{2}\) (since \(m_1 m_2 = -1\)).

The equation of the normal is \(y - 1 = -\frac{1}{2}(x - 2)\), which simplifies to \(2y + x = 4\).

(ii) Solve the simultaneous equations \(2y + x = 4\) and \(y = 5 - \frac{8}{x}\).

Substitute \(y = \frac{4 - x}{2}\) into \(y = 5 - \frac{8}{x}\) to get \(\frac{4 - x}{2} = 5 - \frac{8}{x}\).

Multiply through by \(2x\) to clear fractions: \((4 - x)x = 10x - 16\).

Simplify to \(x^2 + 6x - 16 = 0\). Solving gives \(x = -8\) or \(x = 2\). Since \(x = 2\) is point \(P\), \(Q\) is \((-8, 6)\).

(iii) Use the distance formula to find \(PQ\): \(\sqrt{(2 - (-8))^2 + (1 - 6)^2} = \sqrt{10^2 + 5^2} = \sqrt{125}\).

The length of \(PQ\) is approximately \(11.2\).