(i) To find \(\overrightarrow{OE}\), note that E divides OB in the ratio \(\frac{2}{10}\) since \(OB = \sqrt{8^2 + 6^2} = 10\). Thus, \(\overrightarrow{OE} = \frac{2}{10}(8\mathbf{i} + 6\mathbf{j}) = 1.6\mathbf{i} + 1.2\mathbf{j}\).

(ii) First, find \(\overrightarrow{OD}\) and \(\overrightarrow{BD}\):

\(\overrightarrow{OD} = 1.6\mathbf{i} + 1.2\mathbf{j} + 7\mathbf{k}\)

\(\overrightarrow{BD} = -8\mathbf{i} - 6\mathbf{j} + 1.6\mathbf{i} + 1.2\mathbf{j} + 7\mathbf{k} = -6.4\mathbf{i} - 4.8\mathbf{j} + 7\mathbf{k}\)

Calculate the magnitudes:

\(|\overrightarrow{OD}| = \sqrt{1.6^2 + 1.2^2 + 7^2} = \sqrt{53}\)

\(|\overrightarrow{BD}| = \sqrt{(-6.4)^2 + (-4.8)^2 + 7^2} = \sqrt{113}\)

Find the dot product:

\(\overrightarrow{OD} \cdot \overrightarrow{BD} = 1.6(-6.4) + 1.2(-4.8) + 7(7) = -10.24 - 5.76 + 49 = 33\)

Use the scalar product to find the angle:

\(\cos \angle BDO = \frac{\overrightarrow{OD} \cdot \overrightarrow{BD}}{|\overrightarrow{OD}| \times |\overrightarrow{BD}|} = \frac{33}{\sqrt{53} \times \sqrt{113}}\)

\(\angle BDO = \cos^{-1}\left(\frac{33}{77.4}\right) \approx 64.8^{\circ}\)