(i) To find \(\overrightarrow{OQ}\), note that \(Q\) is the midpoint of \(DF\). Since \(D = (0, 6, 6)\) and \(F = (6, 6, 6)\), the midpoint \(Q\) is \((3, 6, 6)\). Therefore, \(\overrightarrow{OQ} = 3\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}\).

For \(\overrightarrow{PQ}\), since \(P\) is such that \(\overrightarrow{AP} = \frac{1}{3} \overrightarrow{AB}\), and \(A = (6, 0, 0)\), \(B = (6, 6, 0)\), we have \(P = (6, 2, 0)\). Thus, \(\overrightarrow{PQ} = (3 - 6)\mathbf{i} + (6 - 2)\mathbf{j} + (6 - 0)\mathbf{k} = -3\mathbf{i} + \mathbf{j} + 6\mathbf{k}\).

(ii) To find the angle \(OQP\), use the dot product formula: \(\overrightarrow{OQ} \cdot \overrightarrow{PQ} = |\overrightarrow{OQ}| |\overrightarrow{PQ}| \cos \theta\).

Calculate \(\overrightarrow{OQ} \cdot \overrightarrow{PQ} = (3\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) \cdot (-3\mathbf{i} + \mathbf{j} + 6\mathbf{k}) = -9 + 3 + 36 = 30\).

Find magnitudes: \(|\overrightarrow{OQ}| = \sqrt{3^2 + 3^2 + 6^2} = \sqrt{54}\), \(|\overrightarrow{PQ}| = \sqrt{(-3)^2 + 1^2 + 6^2} = \sqrt{46}\).

Thus, \(30 = \sqrt{54} \sqrt{46} \cos \theta\).

\(\cos \theta = \frac{30}{\sqrt{54} \sqrt{46}}\).

\(\theta = \cos^{-1}\left(\frac{30}{\sqrt{54} \sqrt{46}}\right) \approx 53.0^\circ\).