1. Given in the Cartesian coordinate system, $O(0,0)$, $A(6,0)$, $\angle O B A=60^{\circ}$. When the area of $\triangle O A B$ is maximized, the coordinates of point $B$ are $\qquad$
2. $(3,3 \sqrt{3})$ or $(3,-3 \sqrt{3})$.
Since $O$ and $A$ are two fixed points, $\angle O B A=60^{\circ}$, therefore, point $B$ lies on the two major arcs with $O A$ as the chord, and the central angle subtended by chord $O A$ is $120^{\circ}$ (as shown in Figure 7).

Draw $O_{1} C \perp O A$ at point $C$, intersecting the major arc at point $B$. At this moment, the area of $\triangle O A B$ reaches its maximum value.
It is easy to know that $\triangle O A B$ is an equilateral triangle, so $B C=O C \tan 60^{\circ}=3 \sqrt{3}$.
Therefore, the coordinates of point $B$ are $(3,3 \sqrt{3})$ or $(3,-3 \sqrt{3})$.