2.084. $\left(\frac{2-b}{b-1}+2 \cdot \frac{a-1}{a-2}\right):\left(b \cdot \frac{a-1}{b-1}+a \cdot \frac{2-b}{a-2}\right)$;

$a=\sqrt{2}+0.8 ; b=\sqrt{2}-0.2$.
Solution.

$$
\begin{aligned}
& \left(\frac{2-b}{b-1}+2 \cdot \frac{a-1}{a-2}\right):\left(b \cdot \frac{a-1}{b-1}+a \cdot \frac{2-b}{a-2}\right)=\frac{(2-b)(a-2)+2(a-1)(b-1)}{(b-1)(a-2)} \\
& \frac{b(a-1)(a-2)+a(2-b)(b-1)}{(b-1)(a-2)}=\frac{a b-2}{(b-1)(a-2)} \cdot \frac{(b-1)(a-2)}{a^{2} b-a b^{2}-2 a+2 b}= \\
& =\frac{a b-2}{a b(a-b)-2(a-b)}=\frac{a b-2}{(a-b)(a b-2)}=\frac{1}{a-b}=\frac{1}{\sqrt{2}+0.8-\sqrt{2}+0.2}=1 .
\end{aligned}
$$

Answer: 1.