Solution:

Point $X$ is the intersection of circles with diameter $AE$ and $BC$. Thus, there are two possible intersection points. Since $AC \perp BE$, the first point, $X_1$, is the intersection of $AC$ and $BE$, from which we can see $OX_1 = \frac{1}{2}$ as our first answer. Let $X_2$ be the other intersection point.

Let $M$ be the midpoint of $BC$ and $N$ be the midpoint of $AE$. Then $MX_2 = NO = \frac{1}{2}$ and $MO = NX_2 = \frac{\sqrt{3}}{2}$, so $OX_2MN$ is an isosceles trapezoid. By law of cosine on $\triangle OMN$, we have
$$\begin{aligned} MN & = \sqrt{OM^2 + ON^2 - 2 \cdot OM \cdot ON \cos 150^\circ} \\ & = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2}} = \frac{\sqrt{7}}{2} \end{aligned}$$
Moreover, by Ptolemy's theorem,
$$OX_2 \cdot MN = MO^2 - NO^2 = \frac{1}{2}$$
Combining the previous two equations gives $OX_2 = \frac{\sqrt{7}}{7}$.

Solution:
Recall the first paragraph of the previous solution that $X_1 = AC \cap BE$ is the first point. Thus, the second point $X_2$ is the Miquel point of cyclic quadrilateral $ACBE$.

By a well-known property of Miquel point, if $Y = AB \cap CE$, then $Y$ and $X_2$ are inverses with respect to the circumcircle of $ABCDEF$. Thus, $OX_2 \cdot OY = 1$.

One can compute $OY$ as follows: from triangle $BCY$, we get that $BY = BC / \sin 30^\circ = 2$. Thus, by power of point,
$$OY^2 - 1^2 = YB \cdot YA = 2 \cdot 3 = 6 \Longrightarrow OY = \sqrt{7}$$
implying $OX_2 = \frac{\sqrt{7}}{7}$.