(i) To find \(\overrightarrow{CP}\), note that \(C\) is at \((6, 0, 2)\) and \(P\) is at \((0, 6, 0)\). Thus, \(\overrightarrow{CP} = (0 - 6)\mathbf{i} + (6 - 0)\mathbf{j} + (0 - 2)\mathbf{k} = -6\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}\).

For \(\overrightarrow{CQ}\), \(Q\) is at \((0, 6, 5)\). Thus, \(\overrightarrow{CQ} = (0 - 6)\mathbf{i} + (6 - 0)\mathbf{j} + (5 - 2)\mathbf{k} = -6\mathbf{i} + 6\mathbf{j} + 3\mathbf{k}\).

(ii) The scalar product \(\overrightarrow{CP} \cdot \overrightarrow{CQ} = (-6)(-6) + (6)(6) + (-2)(3) = 36 + 36 - 6 = 66\).

The magnitudes are \(|\overrightarrow{CP}| = \sqrt{(-6)^2 + 6^2 + (-2)^2} = \sqrt{76}\) and \(|\overrightarrow{CQ}| = \sqrt{(-6)^2 + 6^2 + 3^2} = \sqrt{81}\).

Using the formula for the angle, \(66 = \sqrt{76} \times \sqrt{81} \times \cos \theta\), we find \(\cos \theta = \frac{66}{\sqrt{76} \times \sqrt{81}}\).

Thus, \(\theta = \cos^{-1}\left(\frac{66}{\sqrt{76} \times \sqrt{81}}\right) \approx 32.7^{\circ}\) or \(0.571 \text{ rad}\).