$$\begin{cases} \log_{2}^{2}(-\log_{2}x) + \log_{2}\log_{2}^{2}x \leq 3 & \\-4 |x^2-1|-3\geq \frac{1}{x^2-1}& \end{cases}$$
What I've tried:
Make substitution $t=x^2-1$ and solve second inequality:
$-4|t|-3 \geq \frac{1}{t} <=> -4|t|-3-\frac{1}{t} \geq 0$
Where I got interval for $t$: $-\frac{1}{4}\leq t < 0$
And then for $x$: $x \in (-1; -\frac{\sqrt{3}}{2}] \cup [\frac{\sqrt{3}}{2}; 1)$
But I have no thoughts how to solve first inequality.
$$ \log_{2}^{2}(-\log_{2}x) + \log_{2}\log_{2}^{2}x \leq 3 $$ $$ \log_{2}^{2}(-\log_{2}x) + 2\log_{2}(\log_{2}x) \leq 3 $$
Determine domain of LHS: $$ -\log_2(x)>0 $$ $$ \log_2(x)<0 $$ $$ \rightarrow x<1 $$
Also: $$ \log_2(x)>0 $$
So domain of this inequality is: $x\in \phi$. Inequality doesn't have any solutions.