(i) To find \(\overrightarrow{AM}\), note that \(M\) is the midpoint of \(AF\). Since \(A = 8\mathbf{i}\) and \(F = 8\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}\), the midpoint \(M\) is:

\(M = \frac{1}{2}(A + F) = \frac{1}{2}(8\mathbf{i} + (8\mathbf{i} + 8\mathbf{j} + 10\mathbf{k})) = \frac{1}{2}(16\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}) = 8\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\)

Thus, \(\overrightarrow{AM} = M - A = (8\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) - 8\mathbf{i} = 4\mathbf{j} + 5\mathbf{k}\).

For \(\overrightarrow{GM}\), since \(G = 3\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}\), we have:

\(\overrightarrow{GM} = M - G = (8\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) - (3\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}) = 5\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}\)

(ii) The scalar product \(\overrightarrow{AM} \cdot \overrightarrow{GM}\) is:

\(\overrightarrow{AM} \cdot \overrightarrow{GM} = (4\mathbf{j} + 5\mathbf{k}) \cdot (5\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}) = 0 \cdot 5 + 4 \cdot (-4) + 5 \cdot (-5) = -16 - 25 = -41\)

The magnitudes are:

\(|\overrightarrow{AM}| = \sqrt{0^2 + 4^2 + 5^2} = \sqrt{41}\)

\(|\overrightarrow{GM}| = \sqrt{5^2 + (-4)^2 + (-5)^2} = \sqrt{66}\)

Using the cosine formula:

\(\overrightarrow{AM} \cdot \overrightarrow{GM} = |\overrightarrow{AM}| \cdot |\overrightarrow{GM}| \cdot \cos \theta\)

\(-41 = \sqrt{41} \cdot \sqrt{66} \cdot \cos \theta\)

\(\cos \theta = \frac{-41}{\sqrt{41} \cdot \sqrt{66}}\)

\(\theta = \cos^{-1}\left(\frac{-41}{\sqrt{41} \cdot \sqrt{66}}\right) \approx 121^{\circ}\)