1. Domain of $f(x)=\log _4 \log _3 \log _7\left(8-\log _2\left(x^2+4 x+5\right)\right)$

For a nested logarithm, each log’s argument must be positive.

Let $A=8-\log_2(x^2+4x+5).$ Then $f(x)=\log_4\big(\log_3(\log_7 A)\big).$

For $\log_4(\log_3(\log_7 A))$ to exist, we need:

• $\log_3(\log_7 A)>0$
• hence $\log_7 A>1$
• hence $A>7$

So, $8-\log_2(x^2+4x+5)>7$ $\Rightarrow \log_2(x^2+4x+5)<1$ $\Rightarrow x^2+4x+5<2.$

Now simplify: $x^2+4x+5=(x+2)^2+1.$ Thus, $(x+2)^2+1<2$ $\Rightarrow (x+2)^2<1$ $\Rightarrow -1<x+2<1$ $\Rightarrow -3<x<-1.$

Therefore the domain of $f$ is $(\alpha,\beta)=(-3,-1).$ So, $\alpha=-3,\quad \beta=-1.$

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

For $\sin^{-1}(t)$ to be defined, $-1\le t\le 1.$ So we need $-1\le \frac{7x+10}{x-2}\le 1, \qquad x\ne 2.$

We solve the two inequalities.

(i) $\dfrac{7x+10}{x-2}\le 1$

$\frac{7x+10}{x-2}-1\le 0$ $\frac{7x+10-(x-2)}{x-2}\le 0$ $\frac{6x+12}{x-2}\le 0$ $\frac{x+2}{x-2}\le 0.$ Critical points: $x=-2,2$. Sign analysis gives $x\in[-2,2).$

(ii) $\dfrac{7x+10}{x-2}\ge -1$

$\frac{7x+10}{x-2}+1\ge 0$ $\frac{7x+10+(x-2)}{x-2}\ge 0$ $\frac{8x+8}{x-2}\ge 0$ $\frac{x+1}{x-2}\ge 0.$ Critical points: $x=-1,2$. Sign analysis gives $x\in(-\infty,-1]\cup(2,\infty).$

Now intersect both results: $[-2,2)\cap\Big(( -\infty,-1]\cup(2,\infty)\Big)=[-2,-1].$

Therefore the domain of $g$ is $[\gamma,\delta]=[-2,-1].$ So, $\gamma=-2,\quad \delta=-1.$

3. Compute $\alpha^2+\beta^2+\gamma^2+\delta^2$

$(-3)^2+(-1)^2+(-2)^2+(-1)^2=9+1+4+1=15.$

Thus, $\alpha^2+\beta^2+\gamma^2+\delta^2=15.$

Hence the correct option is A.