(i) To find \(\overrightarrow{DB}\), note that \(\overrightarrow{OB} = 6\mathbf{i} + 4\mathbf{j}\) and \(\overrightarrow{OD} = 3\mathbf{k}\). Therefore, \(\overrightarrow{DB} = \overrightarrow{OB} - \overrightarrow{OD} = (6\mathbf{i} + 4\mathbf{j}) - 3\mathbf{k} = 6\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}\).

For \(\overrightarrow{DE}\), since \(E\) is the center of the rectangle, \(\overrightarrow{OE} = 3\mathbf{i} + 2\mathbf{j}\). Thus, \(\overrightarrow{DE} = \overrightarrow{OE} - \overrightarrow{OD} = (3\mathbf{i} + 2\mathbf{j}) - 3\mathbf{k} = 3\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\).

(ii) To find angle \(\theta\) between \(\overrightarrow{DB}\) and \(\overrightarrow{DE}\), use the dot product: \(\overrightarrow{DB} \cdot \overrightarrow{DE} = 6 \times 3 + 4 \times 2 + (-3) \times (-3) = 18 + 8 + 9 = 35\).

The magnitudes are \(|\overrightarrow{DB}| = \sqrt{6^2 + 4^2 + (-3)^2} = \sqrt{61}\) and \(|\overrightarrow{DE}| = \sqrt{3^2 + 2^2 + (-3)^2} = \sqrt{22}\).

Using the formula \(\overrightarrow{DB} \cdot \overrightarrow{DE} = |\overrightarrow{DB}| |\overrightarrow{DE}| \cos \theta\), we have:

\(35 = \sqrt{61} \times \sqrt{22} \times \cos \theta\).

Solving for \(\cos \theta\), we get \(\cos \theta = \frac{35}{\sqrt{61} \times \sqrt{22}}\).

Finally, \(\theta = \cos^{-1}\left(\frac{35}{\sqrt{61} \times \sqrt{22}}\right) \approx 17.2^\circ\) (0.300 rad).