(i) To find \(\overrightarrow{AM}\), note that \(M\) is the midpoint of \(AF\). Since \(A = 8\mathbf{i}\) and \(F = 8\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}\), the midpoint \(M\) is:

\(M = \frac{1}{2}(A + F) = \frac{1}{2}(8\mathbf{i} + (8\mathbf{i} + 8\mathbf{j} + 10\mathbf{k})) = \frac{1}{2}(16\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}) = 8\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\)

Thus, \(\overrightarrow{AM} = M - A = (8\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) - 8\mathbf{i} = 4\mathbf{j} + 5\mathbf{k}\).

For \(\overrightarrow{GM}\), since \(G = 3\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}\), we have:

\(\overrightarrow{GM} = M - G = (8\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) - (3\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}) = 5\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}\)

(ii) The scalar product \(\overrightarrow{AM} \cdot \overrightarrow{GM}\) is:

\(\overrightarrow{AM} \cdot \overrightarrow{GM} = (4\mathbf{j} + 5\mathbf{k}) \cdot (5\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}) = 0 \cdot 5 + 4 \cdot (-4) + 5 \cdot (-5) = -16 - 25 = -41\)

The magnitudes are:

\(|\overrightarrow{AM}| = \sqrt{0^2 + 4^2 + 5^2} = \sqrt{41}\)

\(|\overrightarrow{GM}| = \sqrt{5^2 + (-4)^2 + (-5)^2} = \sqrt{66}\)

Using the cosine formula:

\(\overrightarrow{AM} \cdot \overrightarrow{GM} = |\overrightarrow{AM}| \cdot |\overrightarrow{GM}| \cdot \cos \theta\)

\(-41 = \sqrt{41} \cdot \sqrt{66} \cdot \cos \theta\)

\(\cos \theta = \frac{-41}{\sqrt{41} \cdot \sqrt{66}}\)

\(\theta = \cos^{-1}\left(\frac{-41}{\sqrt{41} \cdot \sqrt{66}}\right) \approx 121^{\circ}\)

First, find the vectors BO and BF:

\(\mathbf{BO} = -8\mathbf{i} - 6\mathbf{j}\)

\(\mathbf{BF} = -6\mathbf{j} - 8\mathbf{i} + 7\mathbf{k} + 4\mathbf{i} + 2\mathbf{j} = -4\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\)

Calculate the dot product \(\mathbf{BF} \cdot \mathbf{BO}\):

\(\mathbf{BF} \cdot \mathbf{BO} = (-4)(-8) + (-4)(-6) = 32 + 24 = 56\)

Calculate the magnitudes:

\(|\mathbf{BF}| = \sqrt{(-4)^2 + (-4)^2 + 7^2} = \sqrt{16 + 16 + 49} = \sqrt{81} = 9\)

\(|\mathbf{BO}| = \sqrt{(-8)^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\)

Use the cosine formula:

\(\cos \theta = \frac{\mathbf{BF} \cdot \mathbf{BO}}{|\mathbf{BF}| \times |\mathbf{BO}|} = \frac{56}{9 \times 10} = \frac{56}{90} = \frac{28}{45}\)

Thus, the angle \(\theta = \cos^{-1}\left( \frac{28}{45} \right)\).

First, calculate the vectors \(\overrightarrow{PN}\) and \(\overrightarrow{PM}\).

\(\overrightarrow{PN} = 8\mathbf{i} - 8\mathbf{k}\)

\(\overrightarrow{PM} = 4\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}\)

Calculate the dot product \(\overrightarrow{PN} \cdot \overrightarrow{PM}\):

\(\overrightarrow{PN} \cdot \overrightarrow{PM} = (8)(4) + (0)(4) + (-8)(-6) = 32 + 0 + 48 = 80\)

Calculate the magnitudes:

\(|\overrightarrow{PN}| = \sqrt{8^2 + 0^2 + (-8)^2} = \sqrt{128}\)

\(|\overrightarrow{PM}| = \sqrt{4^2 + 4^2 + (-6)^2} = \sqrt{68}\)

Use the cosine formula:

\(\sqrt{128} \times \sqrt{68} \times \cos M \hat{P} N = 80\)

\(\cos M \hat{P} N = \frac{80}{\sqrt{128} \times \sqrt{68}}\)

Calculate \(M \hat{P} N\):

\(M \hat{P} N = \cos^{-1}\left(\frac{80}{\sqrt{128} \times \sqrt{68}}\right) \approx 31.0^{\circ}\) awrt

(i) To find \(\overrightarrow{DF}\), note that \(D\) is at \((6, 0, 4)\) and \(F\) is at \((0, 0, 6)\). Thus, \(\overrightarrow{DF} = (0 - 6)\mathbf{i} + (0 - 0)\mathbf{j} + (6 - 4)\mathbf{k} = -6\mathbf{i} + 2\mathbf{k}\).

(ii) To find \(\overrightarrow{EF}\), note that \(E\) is at \((6, 3, 4)\) and \(F\) is at \((0, 0, 6)\). Thus, \(\overrightarrow{EF} = (0 - 6)\mathbf{i} + (0 - 3)\mathbf{j} + (6 - 4)\mathbf{k} = -6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}\).

The magnitude of \(\overrightarrow{EF}\) is \(\sqrt{(-6)^2 + (-3)^2 + 2^2} = \sqrt{49} = 7\).

The unit vector in the direction of \(\overrightarrow{EF}\) is \(\frac{1}{7}(-6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k})\).

(iii) To find angle \(EFD\), use the scalar product: \(\overrightarrow{DF} \cdot \overrightarrow{EF} = (-6\mathbf{i} + 2\mathbf{k}) \cdot (-6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}) = 36 + 4 = 40\).

The magnitudes are \(|\overrightarrow{DF}| = \sqrt{40}\) and \(|\overrightarrow{EF}| = 7\).

Thus, \(\cos \angle EFD = \frac{40}{7\sqrt{40}}\).

Therefore, \(\angle EFD = 25.4^\circ\).