(i) To find \(\overrightarrow{OA} \cdot \overrightarrow{OB}\), calculate the dot product:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (2)(7) + (-8)(2) + (4)(-1) = 14 - 16 - 4 = -6\).

Since the dot product is negative, angle AOB is obtuse.

(ii) First, find \(\overrightarrow{AB}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (7\mathbf{i} + 2\mathbf{j} - \mathbf{k}) - (2\mathbf{i} - 8\mathbf{j} + 4\mathbf{k}) = 5\mathbf{i} + 10\mathbf{j} - 5\mathbf{k}\).

Then, \(\overrightarrow{AX} = \frac{2}{5} \overrightarrow{AB} = \frac{2}{5} (5\mathbf{i} + 10\mathbf{j} - 5\mathbf{k}) = 2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\).

Now, find \(\overrightarrow{OX}\):

\(\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = (2\mathbf{i} - 8\mathbf{j} + 4\mathbf{k}) + (2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}) = 4\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}\).

Finally, find the unit vector in the direction of \(\overrightarrow{OX}\):

The magnitude of \(\overrightarrow{OX}\) is \(\sqrt{4^2 + (-4)^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6\).

Thus, the unit vector is \(\frac{1}{6} (4\mathbf{i} - 4\mathbf{j} + 2\mathbf{k})\).