(i) To find \(\cos POQ\), we use the scalar product formula:

\(\overrightarrow{OP} \cdot \overrightarrow{OQ} = (-2)(2) + (3)(1) + (1)(3) = -4 + 3 + 3 = 2\).

The magnitudes are \(|\overrightarrow{OP}| = \sqrt{(-2)^2 + 3^2 + 1^2} = \sqrt{14}\) and \(|\overrightarrow{OQ}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{14}\).

Thus, \(\cos \theta = \frac{\overrightarrow{OP} \cdot \overrightarrow{OQ}}{|\overrightarrow{OP}| |\overrightarrow{OQ}|} = \frac{2}{\sqrt{14} \times \sqrt{14}} = \frac{2}{14} = \frac{1}{7}\).

(ii) The vector \(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \begin{pmatrix} 2 \\ 1 \\ q \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ q-1 \end{pmatrix}\).

The length of \(\overrightarrow{PQ}\) is given by \(\sqrt{4^2 + (-2)^2 + (q-1)^2} = 6\).

Squaring both sides, we have \(16 + 4 + (q-1)^2 = 36\).

Thus, \((q-1)^2 = 16\), giving \(q-1 = 4\) or \(q-1 = -4\).

Therefore, \(q = 5\) or \(q = -3\).