(i) To express \(\overrightarrow{OD}\), note that \(OD\) is a diagonal of the rectangular base \(OABC\) and is elevated by 5 m. The horizontal components are half the lengths of \(OA\) and \(OC\), so \(\overrightarrow{OD} = 4\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\).

The magnitude of \(\overrightarrow{OD}\) is calculated using the Pythagorean theorem: \(\sqrt{4^2 + 4^2 + 5^2} = \sqrt{57} \approx 7.55 \, \text{m}\).

(ii) Vector \(\overrightarrow{OB} = 14\mathbf{i} + 8\mathbf{j}\).

The dot product \(\overrightarrow{OD} \cdot \overrightarrow{OB} = 4 \times 14 + 4 \times 8 = 88\).

The magnitudes are \(\sqrt{57}\) for \(\overrightarrow{OD}\) and \(\sqrt{260}\) for \(\overrightarrow{OB}\).

Using the dot product formula: \(88 = \sqrt{57} \times \sqrt{260} \times \cos \theta\).

Solving for \(\theta\), \(\cos \theta = \frac{88}{\sqrt{57} \times \sqrt{260}}\), giving \(\theta \approx 43.7^\circ\).

(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:

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

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

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

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

(i) The vector \(\overrightarrow{MO}\) is expressed as \(4\mathbf{i} - 6\mathbf{k}\).

The vector \(\overrightarrow{MC}\) is expressed as \(4\mathbf{i} + 4\mathbf{j} + 6\mathbf{k}\).

(ii) To find the angle \(\angle OMC\), we use the dot product formula:

\(\overrightarrow{MO} \cdot \overrightarrow{MC} = 4 \times 4 + 0 \times 4 + (-6) \times 6 = 16 - 36 = -20\).

The magnitudes are \(|\overrightarrow{MO}| = \sqrt{4^2 + 6^2} = \sqrt{52}\) and \(|\overrightarrow{MC}| = \sqrt{4^2 + 4^2 + 6^2} = \sqrt{68}\).

Using the cosine formula: \(\overrightarrow{MO} \cdot \overrightarrow{MC} = |\overrightarrow{MO}| |\overrightarrow{MC}| \cos \theta\).

\(-20 = \sqrt{52} \times \sqrt{68} \times \cos \theta\).

\(\cos \theta = \frac{-20}{\sqrt{52} \times \sqrt{68}}\).

\(\theta = \cos^{-1}\left(\frac{-20}{\sqrt{52} \times \sqrt{68}}\right) \approx 109.7^\circ\).

(i) To show that \(AB\) is perpendicular to \(BC\), calculate the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 9 \end{pmatrix}\)

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\)

Calculate the dot product:

\(\overrightarrow{AB} \cdot \overrightarrow{BC} = 3 \times 2 + (-6) \times 1 + 9 \times 0 = 6 - 6 = 0\)

Since the dot product is zero, \(AB\) is perpendicular to \(BC\).

(ii) To show that \(ABCD\) is a trapezium, calculate \(\overrightarrow{DC}\):

\(\overrightarrow{DC} = \overrightarrow{OC} - \overrightarrow{OD} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ 6 \end{pmatrix}\)

Check if \(\overrightarrow{AB}\) is parallel to \(\overrightarrow{DC}\):

\(\overrightarrow{AB} = k \overrightarrow{DC}\) where \(k = \frac{3}{2}\)

Since \(AB \parallel DC\), \(ABCD\) is a trapezium.

(iii) To find the area of \(ABCD\), calculate the magnitudes:

\(|\overrightarrow{AB}| = \sqrt{9 + 36 + 81} = \sqrt{126} = 11.22\)

\(|\overrightarrow{DC}| = \sqrt{4 + 16 + 36} = \sqrt{56} = 7.483\)

\(|\overrightarrow{BC}| = \sqrt{4 + 1 + 0} = \sqrt{5} = 2.236\)

Area = \(\frac{1}{2} (|\overrightarrow{AB}| + |\overrightarrow{DC}|) \times |\overrightarrow{BC}| = \frac{1}{2} (11.22 + 7.483) \times 2.236 = 20.92\)

(i) To find \(\overrightarrow{MF}\), note that M is the midpoint of AB, so M is at \((4, 0, 0)\). F is directly above D at \((0, 4, 7)\). Thus, \(\overrightarrow{MF} = (0 - 4)\mathbf{i} + (4 - 0)\mathbf{j} + (7 - 0)\mathbf{k} = -4\mathbf{i} + 2\mathbf{j} + 7\mathbf{k}\).

(ii) N is the midpoint of FE. Since F is at \((0, 4, 7)\) and E is at \((8, 0, 0)\), N is at \((4, 2, 3.5)\). Thus, \(\overrightarrow{FN} = (4 - 0)\mathbf{i} + (2 - 4)\mathbf{j} = 2\mathbf{i} - \mathbf{j}\).

(iii) \(\overrightarrow{MN} = \overrightarrow{MF} + \overrightarrow{FN} = (-4\mathbf{i} + 2\mathbf{j} + 7\mathbf{k}) + (2\mathbf{i} - \mathbf{j}) = -2\mathbf{i} + \mathbf{j} + 7\mathbf{k}\).

(iv) To find angle FMN, use the scalar product: \(\overrightarrow{MF} \cdot \overrightarrow{MN} = -4 \times -2 + 2 \times 1 + 7 \times 7 = 59\).

The magnitudes are \(|\overrightarrow{MF}| = \sqrt{(-4)^2 + 2^2 + 7^2} = \sqrt{69}\) and \(|\overrightarrow{MN}| = \sqrt{(-2)^2 + 1^2 + 7^2} = \sqrt{54}\).

Thus, \(\cos FMN = \frac{59}{\sqrt{69} \times \sqrt{54}} = \frac{59}{\sqrt{3726}}\).

Therefore, \(FMN = \cos^{-1}\left(\frac{59}{\sqrt{3726}}\right) \approx 14.9^\circ \text{ or } 0.259 \text{ radians}\).