12. Let $x, y \in \mathbf{R}$, if $2 x, 1, y-1$ form an arithmetic sequence, and $y+3,|x+1|+|x-1|$, $\cos \left(\arcsin \sqrt{1-x^{2}}\right)$ form a geometric sequence, then $(x+1)(y+1)=$ $\qquad$ .
12. 4 or $2(\sqrt{17}-3)$ Given $-1 \leqslant x \leqslant 1$, and $|x| \cdot(y+3)=4$ as well as $2 x+y=3$.
（1）When $x \geqslant 0$, solving yields $x=1, y=1$, hence $(x+1)(y+1)=4$;
（2）When $x<0$, solving yields $x=\frac{3-\sqrt{17}}{2}, y=\sqrt{17}$, hence $(x+1)(y+1)=2(\sqrt{17}-3)$.