Explanation

(a)
Given that, $\overrightarrow{\mathbf{a}} = (1,1,4) = \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}4\widehat{\mathbf{k}}$
and $\overrightarrow{\mathbf{b}} = (1, - 1,\ 4) = \widehat{\mathbf{i}}\mathbf{-}\widehat{\mathbf{j}}\mathbf{+ 4}\widehat{k}$
$\therefore\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}\mathbf{=}2\widehat{\mathbf{i}}\mathbf{+}8\widehat{\mathbf{k}}$
$\Rightarrow \overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}}\mathbf{=}2\widehat{\mathbf{j}}$
Let $\theta$ be the angle between $\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}}$, then
$\cos{\theta = \frac{\left( \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right) \bullet (\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}})}{\left| \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right|\mathbf{|}\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}}|\ }}$
$= \frac{\left( 2\widehat{\mathbf{i}} + 0\widehat{\mathbf{j}} + 8\widehat{\mathbf{k}} \right) \bullet (0\widehat{\mathbf{i}} + 2\widehat{\mathbf{j}} + 0\widehat{\mathbf{k}})}{\sqrt{2^{2} + 0^{2} + 8^{2}}\sqrt{0^{2} + 2^{2} + 0^{2}}}$
$= \frac{0 + 0 + 0}{\sqrt{4 + 64}\sqrt{4}} = 0$
$\Rightarrow \cos{\theta = \cos{\theta{^\circ}\ \Rightarrow \ \theta = \frac{\pi}{2} = 90{^\circ}}}$