(i) Since D is the midpoint of AB, the coordinates of D are \((4, 3, 0)\) because A is at \((8, 0, 0)\) and B is at \((0, 6, 0)\). Thus, \(\overrightarrow{OD} = 4\mathbf{i} + 3\mathbf{j}\).

For \(\overrightarrow{CD}\), since C is at \((0, 0, 10)\), \(\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} = 4\mathbf{i} + 3\mathbf{j} - 10\mathbf{k}\).

(ii) To find \(\angle ODC\), use the scalar product formula:

\(\overrightarrow{OD} \cdot \overrightarrow{CD} = |\overrightarrow{OD}| \times |\overrightarrow{CD}| \times \cos \theta\)

\(\overrightarrow{OD} \cdot \overrightarrow{CD} = (4)(4) + (3)(3) + (0)(-10) = 16 + 9 = 25\)

The magnitudes are \(|\overrightarrow{OD}| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5\) and \(|\overrightarrow{CD}| = \sqrt{4^2 + 3^2 + (-10)^2} = \sqrt{125}\).

Thus, \(25 = 5 \times \sqrt{125} \times \cos \theta\).

Solving for \(\cos \theta\), we get \(\cos \theta = \frac{25}{5 \times \sqrt{125}} = \frac{1}{\sqrt{5}}\).

Therefore, \(\theta = \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) \approx 63.4^\circ\) or 1.11 radians.