Explanation

(a)
Since, $\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{c}}\ \Longrightarrow (\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}})^{2} = {\overrightarrow{\mathbf{c}}}^{2}$
$\Longrightarrow \ \left| \overrightarrow{\mathbf{a}} \right|^{2} + \left| \overrightarrow{\mathbf{b}} \right|^{2} + 2\left| \overrightarrow{\mathbf{a}} \right|\left| \overrightarrow{\mathbf{b}} \right|\cos{\theta = \left| \overrightarrow{\mathbf{c}} \right|^{2}}$
$\Longrightarrow 2(1 + \cos{\theta) = 1\ \Longrightarrow \cos\theta = - \frac{1}{2}\ \ \ \ \lbrack\therefore\left| \overrightarrow{\mathbf{a}} \right| = \left| \overrightarrow{\mathbf{b}} \right| = \left| \overrightarrow{\mathbf{c}} \right| = 1,\ given\rbrack}$
Now, $\left| \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^{2} = \left| \overrightarrow{\mathbf{a}}|^{2} + {|\overrightarrow{\mathbf{b|}}}^{2} - 2 \right|\overrightarrow{\mathbf{a}} \right|$|$\overrightarrow{\mathbf{b}}|\cos\theta$
$= 1 + 1 + 2 \bullet \frac{1}{2} = 3$
$\Longrightarrow \left| \overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}} \right| = \sqrt{3}$