2. Let $\angle X O Y=30^{\circ}$, point $A$ is inside this angle, draw $A B \perp O X, A C \perp O Y ; B, C$ are the feet of the perpendiculars. If $A C=3, O B=8$, then the length of $A O$ is $\qquad$ .
2. $\frac{14 \sqrt{3}}{3}$.

As shown in Figure 1, with $O$ as the origin and $OX$ as the positive direction of the $x$-axis, a Cartesian coordinate system is established. Let the coordinates of point $A$ be $\left(x_{A}, y_{A}\right)$, where $x_{A}=OB=8$. Clearly, the equation of the line $OY$ is
$$
y=x \tan 30^{\circ}=\frac{\sqrt{3}}{3} x .
$$

Draw a line $l \parallel OY$ through point $A$, and let $l$ intersect the $y$-axis at point $D$. Draw $OE \perp l$, with $E$ as the foot of the perpendicular. Clearly, $OE=AC=3$, and $\angle DOE=30^{\circ}$.
Therefore, $OD=\frac{OE}{\cos 30^{\circ}}=2 \sqrt{3}$, so $D(0,-2 \sqrt{3})$.
Thus, the equation of the line $l$ is $y=\frac{\sqrt{3}}{3} x-2 \sqrt{3}$. Hence, $y_{A}=\frac{\sqrt{3}}{3} x_{A}-2 \sqrt{3}=\frac{\sqrt{3}}{3} \times 8-2 \sqrt{3}=\frac{2 \sqrt{3}}{3}$.

Therefore, $OA=\sqrt{x_{A}^{2}+y_{A}^{2}}=\sqrt{8^{2}+\left(\frac{2 \sqrt{3}}{3}\right)^{2}}=\frac{14 \sqrt{3}}{3}$.