(i) For \(\mathbf{u}\) to be perpendicular to \(\mathbf{v}\), their dot product must be zero:

\(\mathbf{u} \cdot \mathbf{v} = q \cdot 8 + 2 \cdot (q-1) + 6 \cdot (q^2-7) = 0\)

\(8q + 2q - 2 + 6q^2 - 42 = 0\)

\(6q^2 + 10q - 44 = 0\)

Solving this quadratic equation gives \(q = 2\) and \(q = -\frac{11}{3}\).

(ii) When \(q = 0\), \(\mathbf{u} = \begin{pmatrix} 0 \\ 2 \\ 6 \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} 8 \\ -1 \\ -7 \end{pmatrix}\).

The dot product \(\mathbf{u} \cdot \mathbf{v} = 0 \cdot 8 + 2 \cdot (-1) + 6 \cdot (-7) = -2 - 42 = -44\).

The magnitudes are \(|\mathbf{u}| = \sqrt{0^2 + 2^2 + 6^2} = \sqrt{40}\) and \(|\mathbf{v}| = \sqrt{8^2 + (-1)^2 + (-7)^2} = \sqrt{114}\).

\(\cos \theta = \frac{-44}{\sqrt{40} \times \sqrt{114}} = \frac{-44}{\sqrt{4560}} = \frac{-4}{\sqrt{11}}\).

\(\theta = \cos^{-1}\left(\frac{-4}{\sqrt{11}}\right) \approx 130.7^\circ\) or \(2.28 \text{ radians}\).