Explanation

(b)
Since, $\left| \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} \right| = \sqrt{3}$
$\Longrightarrow \left| \overrightarrow{\mathbf{a}} \right|^{2} + |{\overrightarrow{\mathbf{b|}}}^{2} + 2\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} = 3$
$\Longrightarrow \ \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} = \frac{1}{2}\ldots..(i)$
$\because\lbrack\left| \overrightarrow{\mathbf{a}} \right| = \left| \overrightarrow{\mathbf{b}} \right| = 1,\ given\rbrack$
$\therefore\left( 3\overrightarrow{\mathbf{a}} - 4\overrightarrow{\mathbf{b}} \right) \bullet \left( 2\overrightarrow{\mathbf{a}} + 5\overrightarrow{\mathbf{b}} \right) = 6 + 7\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} - 20$
$= 6 + \frac{7}{2} - 20$
$= - \frac{21}{2}\ \ \ \ \lbrack from\ Eq.(i)\rbrack$