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Solution

For $f(x)=\log_{4}\big(\log_{3}\big(\log_{7}\big(8-\log_{2}(x^{2}+4x+5)\big)\big)\big)$

Condition:
$\log_{3}(\cdot)>0 \Rightarrow \log_{7}\big(8-\log_{2}(x^{2}+4x+5)\big)>1$

$\Rightarrow 8-\log_{2}(x^{2}+4x+5)>7$
$\Rightarrow \log_{2}(x^{2}+4x+5)<1$
$\Rightarrow x^{2}+4x+5<2$
$\Rightarrow (x+2)^2<1$
$\Rightarrow -3$
So $(\alpha,\beta)=(-3,-1)$

For $g(x)=\sin^{-1}\left(\frac{7x+10}{x-2}\right)$

Condition: $-1\le \frac{7x+10}{x-2}\le 1$

Solve:
$\frac{7x+10}{x-2}\le 1 \Rightarrow \frac{6x+12}{x-2}\le 0 \Rightarrow \frac{x+2}{x-2}\le 0$
$\Rightarrow -2\le x<2$

$\frac{7x+10}{x-2}\ge -1 \Rightarrow \frac{8x+8}{x-2}\ge 0 \Rightarrow \frac{x+1}{x-2}\ge 0$
$\Rightarrow x\le -1 \text{ or } x>2$

Intersection: $[-2,-1]$
So $(\gamma,\delta)=(-2,-1)$

Now:
$\alpha^2+\beta^2+\gamma^2+\delta^2=9+1+4+1=15$

$\boxed{15}$