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