1. Let $x, y \in \mathbf{R}$, if $2 x, 1, y-1$ form an arithmetic sequence, and $y+3,|x+1|+|x-1|, \cos (\arccos x)$ form a geometric sequence, then the value of $(x+1)(y+1)$ is $\qquad$.
1. 4 .

From $2 x, 1, y-1$ forming an arithmetic sequence, we get
$$
y=3-2 x .
$$

From $\cos (\arccos x)=x$ and $-1 \leqslant x \leqslant 1$, we deduce $|x+1|+|x-1|=2$. Therefore, $y+3,2, x$ form a geometric sequence, which gives
$$
x(y+3)=4 \text {. }
$$

Substituting equation (1) into equation (2), we get $2 x^{2}-6 x+4=0$. Solving this, we find $x_{1}=2$ (discard), $x_{2}=1$. Thus, $y=1$. Hence, $(x+1)(y+1)=4$.