Explanation

(b) $a^{2} + b^{2} + 2ab\ cos\theta$
$= - a^{2} + b^{2} - 2ab\ cos\theta$
or $4ab\ cos\theta = 0$
But $4ab \neq 0\therefore cos\theta = 0$ or $\theta = 90{^\circ}$
Again
$\left| \overrightarrow{a} + \overrightarrow{b} \right|$ and $\left| \overrightarrow{a} - \overrightarrow{b} \right|$ are the diagonals of parallelogram whose adjacent sides are $\overrightarrow{a}$ and $\overrightarrow{b}$.
Since $\left| \overrightarrow{a} + \overrightarrow{b} \right| = \left| \overrightarrow{a} - \overrightarrow{b} \right|$, therefore, the two diagonals of a parallelogram are equal. So, think of square. This leads to $\theta = 90{^\circ}$.