(i) To find \(\overrightarrow{AM}\), note that \(M\) is the midpoint of \(OX\). Since \(OX = 10\mathbf{k}\), \(M\) is at \(5\mathbf{k}\). \(A\) is at \(8\mathbf{i}\). Thus, \(\overrightarrow{AM} = 5\mathbf{k} - 8\mathbf{i} = -6\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}\).

For \(\overrightarrow{AC}\), \(C\) is at \(8\mathbf{j}\). Thus, \(\overrightarrow{AC} = 8\mathbf{j} - 8\mathbf{i} = -8\mathbf{i} + 8\mathbf{j}\).

(ii) The scalar product \(\overrightarrow{AM} \cdot \overrightarrow{AC} = (-6)(-8) + (2)(8) + (5)(0) = 48 + 16 = 64\).

The magnitudes are \(|\overrightarrow{AM}| = \sqrt{(-6)^2 + 2^2 + 5^2} = \sqrt{65}\) and \(|\overrightarrow{AC}| = \sqrt{(-8)^2 + 8^2} = \sqrt{128}\).

Using the cosine formula, \(64 = \sqrt{128} \sqrt{65} \cos \theta\).

Solving for \(\cos \theta\), we find \(\theta = 45.4^\circ\).