If f(x) = \ cases x + a, & x ≤ 0 |x - 4|, & x > 0 cases ) and g(x)...

Given f\left(x\right) = \left\{\begin{cases}x + a; & x \le 0 \\ \left|x - 4\right|; & x > 0\end{cases}\right)and g\left(x\right) = \left\{$\begin{cases}x + 1; & x < 0 \\ {\left(x - 4\right)}^{2} + b; & x \ge 0\end{cases}$\right)

Given $f\left(x\right) \text{ and } g\left(x\right)$ are continuous on $R$

So, for continuity

$f\left({0}^{-}\right) = f\left(0\right) = f\left({0}^{+}\right)$

$⇒ a = \left|0 - 4\right|$

$⇒ a = 4$

And $g\left({0}^{-}\right) = g\left(0\right) = g\left({0}^{+}\right)$

$⇒ 1 = {\left(0 - 4\right)}^{2} + b$

$⇒ b = 1 - 16 = - 15$

Then, $g\left(f\left(2\right)\right) + f\left(g\left(- 2\right)\right)$

$= g\left(2\right) + f\left(- 1\right)$

$= {\left(2 - 4\right)}^{2} - 15 + \left(- 1 + 4\right)$

$= 4 - 15 + 3 = - 8$