2. When $b=-1$, the equation is reduced to $|x|-\arcsin x-\arccos x-|x|+1+a=0$, which for $x \in[-1 ; 1]$ is equivalent to $1+a-\frac{\pi}{2}=0$. Thus, when $b=-1$, a solution exists only when $a=\frac{\pi}{2}-1$.

On the other hand, when $a=\frac{\pi}{2}-1$, the equation

$$
|x|-\arcsin x+b \cdot(\arccos x+|x|-1)+\frac{\pi}{2}-1=0
$$

has a root $x=1$ for any value of $b$.
Answer: $\frac{\pi}{2}-1$.

Answer to option: $1-2: \frac{\pi}{2}-2$.