$8.26 \quad \log _{0.2}^{2}(x-1)^{2}>4$.

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$8.26 \quad \log _{0.2}^{2}(x-1)^{2}>4$.
8.26 The given inequality can be rewritten as $\left|\log _{0.2}(x-1)\right|>2$, from which we have either.

$\log _{0.2}(x-1) > 2$ or $\log _{0.2}(x-1) < -2$.

Solving inequality (1). Since $-2=\log _{0.2} 0.2^{-2}$, taking into account that $0.2^2 = 0.04$, we get $x-1 < 0.04$, i.e., $x < 1.04$. Solving inequality (2), we find $0 < x-1 < 0.2^{-2}$, i.e., $x > 26$. Solving inequality (2), we find $0 < x-1 < 0.04$, i.e., $1 < x < 1.04$.

Answer: $(1 ; 1.04) \cup(26, \infty)$.