(i) To find \(\overrightarrow{CE}\), we use the coordinates of C and E. Since E is 8 units above D, and D is at (0, 2, 0), E is at (0, 2, 8). C is at (4, 1, 0). Thus, \(\overrightarrow{CE} = (0 - 4)\mathbf{i} + (2 - 1)\mathbf{j} + (8 - 0)\mathbf{k} = -4\mathbf{i} - \mathbf{j} + 8\mathbf{k}\).

The length of \(CE\) is \(|CE| = \sqrt{(-4)^2 + (-1)^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9\).

(ii) To find angle ECA, we first find \(\overrightarrow{CA}\). A is at (7, 0, 0), so \(\overrightarrow{CA} = (7 - 4)\mathbf{i} + (0 - 1)\mathbf{j} = 3\mathbf{i} - \mathbf{j}\).

The scalar product \(\overrightarrow{CE} \cdot \overrightarrow{CA} = (-4\mathbf{i} - \mathbf{j} + 8\mathbf{k}) \cdot (3\mathbf{i} - \mathbf{j}) = -12 + 1 = -11\).

The magnitudes are \(|\overrightarrow{CE}| = 9\) and \(|\overrightarrow{CA}| = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}\).

Thus, \(\cos \theta = \frac{-11}{9\sqrt{10}}\), and \(\theta = \cos^{-1} \left( \frac{-11}{9\sqrt{10}} \right)\).

Rewriting in the required form, \(\cos^{-1} \left( \frac{-1}{\sqrt{18}} \right)\).