1. Given that the vertices $A$ and $C$ of $\triangle A B C$ lie on the graph of the inverse proportion function $y=\frac{\sqrt{3}}{x}(x>0)$, $\angle A C B=90^{\circ}$, $\angle A B C=30^{\circ}$, $A B \perp x$-axis, point $B$ is above point $A$, and $A B=6$. Then the coordinates of point $C$ are
二, 1. $\left(\frac{\sqrt{3}}{2}, 2\right)$.
Draw $C D \perp A B$ at point $D$.
Let $C\left(m, \frac{\sqrt{3}}{m}\right), A\left(n, \frac{\sqrt{3}}{n}\right)$.
From the problem, we know $n>m>0, D\left(n, \frac{\sqrt{3}}{m}\right)$.

Thus, point $C\left(\frac{\sqrt{3}}{2}, 2\right)$.