2. 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 $\qquad$ .
2.4.

Given that $2 x, 1, y-1$ form an arithmetic sequence, we have
$$
y=3-2 x \text{. }
$$

From $\cos (\arccos x)=x$ and $-1 \leqslant x \leqslant 1$, it follows that
$$
|x+1|+|x-1|=2 \text{. }
$$

Thus, $y+3, 2, x$ form a geometric sequence, which gives
$$
x(y+3)=4 \text{. }
$$

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