(2) If the set $A=\{x \mid \sqrt{x-3}=a x+1, x \in \mathbf{R}\}$ is an empty set, then the range of real number $a$ is $\qquad$.
(2) $\left(-\infty,-\frac{1}{3}\right) \cup\left(\frac{1}{6},+\infty\right)$