Explanation

(b)
Let $\overrightarrow{\mathbf{AB}} = \overrightarrow{\mathbf{a}} = 3\ \overrightarrow{\alpha} - \overrightarrow{\beta},\ \overrightarrow{\mathbf{BC}} = \overrightarrow{\mathbf{b}} = \overrightarrow{\alpha} + 3\overrightarrow{\beta}$
Diagonal $\overrightarrow{\mathbf{AC}} = \overrightarrow{\mathbf{AB}} + \overrightarrow{\mathbf{BC}} = \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}$
$\mathbf{\Longrightarrow}\left| \overrightarrow{\mathbf{AC}} \right| = \left| \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} \right|$
$\mathbf{\Longrightarrow}\left| \overrightarrow{\mathbf{AC}} \right| = \left| 4\ \overrightarrow{\alpha} + 2\overrightarrow{\beta} \right|$
$\mathbf{\Longrightarrow}\left| \overrightarrow{\mathbf{AC}} \right|^{\mathbf{2}}\mathbf{=}16{\overrightarrow{\alpha}}^{2} + 4{\overrightarrow{\beta}}^{2} + 16\overrightarrow{\alpha} \bullet \overrightarrow{\beta}$
$\mathbf{\Longrightarrow}\left| \overrightarrow{\mathbf{AC}} \right|^{\mathbf{2}} = 64 + 16 + 16\left| \overrightarrow{\alpha} \right|\left| \overrightarrow{\beta} \right|\cos\frac{\pi}{3}$
$\mathbf{\Longrightarrow}\left| \overrightarrow{\mathbf{AC}} \right|^{\mathbf{2}} = 80 + 16 \times 4 \times \frac{1}{2} = 112$
$\Longrightarrow \ \left| \overrightarrow{\mathbf{AC}} \right| = 4\sqrt{7}$
Other diagonal is $|\overrightarrow{\mathbf{BD}} = |\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}|$
$\mathbf{\Longrightarrow}\left| \overrightarrow{\mathbf{BD}} \right|^{\mathbf{2}}\mathbf{= |}2\overrightarrow{\alpha} - 4\overrightarrow{\beta}|^{2}$
$= 4\left| \overrightarrow{\alpha}|^{2} + 16 \right|\overrightarrow{\beta}|^{2} - 16\ \left| \overrightarrow{\alpha} \right|\left| \overrightarrow{\beta} \right|\cos\frac{\pi}{3}$
$= 64 + 16 - 16 \times 4 \times \frac{1}{2} = 48$
$\Longrightarrow |\overrightarrow{\mathbf{BD}} = \sqrt{48} = 4\sqrt{3}$