(i) At \(x = 1\), the gradient \(m\) of the tangent is \(2\). The gradient of the normal is \(-\frac{1}{2}\) since \(m \cdot m_\text{normal} = -1\).

The equation of the normal is \(y - 8 = -\frac{1}{2}(x - 1)\), which simplifies to \(2y + x = 17\).

The normal meets the x-axis when \(y = 0\), giving \(x = 17\), so \(R(17, 0)\).

The normal meets the y-axis when \(x = 0\), giving \(y = 8.5\), so \(Q(0, 8.5)\).

The mid-point of \(QR\) is \(\left( \frac{0 + 17}{2}, \frac{8.5 + 0}{2} \right) = \left( \frac{17}{2}, \frac{9}{2} \right)\).

(ii) Integrate \(\frac{dy}{dx} = \frac{4}{\sqrt{6 - 2x}}\) to find \(y\).

\(y = \int \frac{4}{\sqrt{6 - 2x}} \, dx = 4(6 - 2x)^{\frac{1}{2}} \cdot \left(-\frac{1}{2}\right) + c\).

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

Substitute \(P(1, 8)\) into the equation: \(8 = \frac{1}{2}(1) - 2 + c\).

Solving gives \(c = 16\).

Thus, the equation of the curve is \(y = \frac{1}{2}x - 2 + 16\).