14. (2004 National College Entrance Examination - Liaoning Paper) Let the universal set $U=\mathbf{R}$.
(1) Solve the inequality for $x$: $|x-1|+a-1 \geqslant 0(a \in \mathbf{R})$
(2) Let $A$ be the solution set of the inequality in (1). The set $B=\left\{x \left\lvert\, \sin \left(\pi x-\frac{\pi}{3}\right)+\sqrt{3} \cos \left(\pi x-\frac{\pi}{3}\right)=0\right.\right\}$, if $\left(C_{\cup} A\right) \cap B$ has exactly three elements, find the range of values for $a$.
14. (1) When $a>1$, the solution set is $R$; when $a \leqslant 1$, the solution set is $\{x \mid x2-a\}$. (2) When $a>1$, $C_{U} A=\Phi$; when $a \leqslant 1$, $C_{\mathrm{U}} A=\{x \mid a \leqslant x \leqslant 2-a\}$

Since $\sin \left(\pi x-\frac{\pi}{3}\right)+\sqrt{3} \cos \left(\pi x-\frac{\pi}{3}\right)=2 \sin \pi x$. From $\sin \pi x=0$ we get: $x=R(R \in \mathbf{Z})$. Therefore, $B=\mathbf{Z}$. When $\left(C_{U} A\right) \cap B$ contains three elements, $a$ should satisfy $\left\{\begin{array}{l}a<1 \\ 2 \leqslant 2-a<3 \Rightarrow-1<a \leqslant 0 \\ -1<a \leqslant 0\end{array}\right.$