(i) To find the unit vector in the direction of \(\overrightarrow{PM}\), we first calculate \(\overrightarrow{PM}\) as follows:

\(\overrightarrow{PM} = 2\mathbf{i} - 10\mathbf{k} + \frac{1}{2}(6\mathbf{j} + 8\mathbf{k})\)

Simplifying, \(\overrightarrow{PM} = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\).

The magnitude of \(\overrightarrow{PM}\) is \(\sqrt{4 + 9 + 36} = 7\).

Thus, the unit vector is \(\frac{1}{7} (2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k})\).

(ii) For \(\angle ATP = \cos^{-1}\left(\frac{2}{7}\right)\), we use the vectors:

\(\overrightarrow{AT} = 6\mathbf{j} + 8\mathbf{k}\)

\(\overrightarrow{PT} = a\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}\)

\(\cos(ATP) = \frac{(6\mathbf{j} + 8\mathbf{k}) \cdot (a\mathbf{i} + 6\mathbf{j} - 2\mathbf{k})}{\sqrt{36 + 64}\sqrt{a^2 + 36 + 4}}\)

\(= \frac{36 - 16}{\sqrt{100}\sqrt{a^2 + 40}} = \frac{20}{10\sqrt{a^2 + 40}}\)

\(\frac{2}{\sqrt{a^2 + 40}} = \frac{2}{7}\)

Solving, \(a = 3\).