(i) To express \(\overrightarrow{OD}\), note that \(OD\) is a diagonal of the rectangular base \(OABC\) and is elevated by 5 m. The horizontal components are half the lengths of \(OA\) and \(OC\), so \(\overrightarrow{OD} = 4\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\).

The magnitude of \(\overrightarrow{OD}\) is calculated using the Pythagorean theorem: \(\sqrt{4^2 + 4^2 + 5^2} = \sqrt{57} \approx 7.55 \, \text{m}\).

(ii) Vector \(\overrightarrow{OB} = 14\mathbf{i} + 8\mathbf{j}\).

The dot product \(\overrightarrow{OD} \cdot \overrightarrow{OB} = 4 \times 14 + 4 \times 8 = 88\).

The magnitudes are \(\sqrt{57}\) for \(\overrightarrow{OD}\) and \(\sqrt{260}\) for \(\overrightarrow{OB}\).

Using the dot product formula: \(88 = \sqrt{57} \times \sqrt{260} \times \cos \theta\).

Solving for \(\theta\), \(\cos \theta = \frac{88}{\sqrt{57} \times \sqrt{260}}\), giving \(\theta \approx 43.7^\circ\).