(i) To find the length of OB, use the Pythagorean theorem in triangle OBC. Since O is the midpoint of AC, the height OB is perpendicular to AC. Given AB = 5 units and BC = 5 units, the height OB can be calculated as:

\(OB = \sqrt{AB^2 - \left(\frac{AC}{2}\right)^2} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4\)

(ii) To express \(\overrightarrow{MC}\) and \(\overrightarrow{MN}\) in terms of i, j, and k:

Since M is the midpoint of BE, and N is the midpoint of DF, we have:

\(\overrightarrow{MC} = 3\mathbf{i} - 6\mathbf{j} - 4\mathbf{k}\)

\(\overrightarrow{MN} = 6\mathbf{j} - 4\mathbf{k}\)

(iii) To evaluate \(\overrightarrow{MC} \cdot \overrightarrow{MN}\) and find angle CMN:

Calculate the dot product:

\(\overrightarrow{MC} \cdot \overrightarrow{MN} = (3)(0) + (-6)(6) + (-4)(-4) = -36 + 16 = -20\)

Find the magnitudes:

\(|\overrightarrow{MC}| = \sqrt{3^2 + (-6)^2 + (-4)^2} = \sqrt{61}\)

\(|\overrightarrow{MN}| = \sqrt{0^2 + 6^2 + (-4)^2} = \sqrt{52}\)

Use the dot product formula to find the angle:

\(\overrightarrow{MC} \cdot \overrightarrow{MN} = |\overrightarrow{MC}| |\overrightarrow{MN}| \cos \theta\)

\(-20 = \sqrt{61} \sqrt{52} \cos \theta\)

\(\cos \theta = \frac{-20}{\sqrt{61} \sqrt{52}}\)

\(\theta = \cos^{-1}\left(\frac{-20}{\sqrt{61} \sqrt{52}}\right) \approx 111^\circ\)