Solution:

Answer: $\frac{1+\sqrt{13}}{2}$

Inspection shows that one point must be on the semicircle and the other must be on the side of the hexagon directly opposite the edge with the semicircle, the bottom edge of the hexagon in the above diagram. Let $O$ be the center of the semicircle and let $M$ be the midpoint of the bottom edge.

We will determine the longest distance between points in the figure by comparing the lengths of all the segments with one endpoint on the bottom edge and the other endpoint on the semicircle. Fix a point $A$ on the bottom edge of the hexagon. Suppose that $B$ is chosen on the semicircle such that $AB$ is as long as possible. Let $C$ be the circle centered at $A$ with radius $AB$. If $C$ is not tangent to the semicircle, then part of the semicircle is outside $C$, so we could pick a $B'$ on the semicircle such that $AB'$ is longer than $AB$. So $C$ must be tangent to the semicircle, and $AB$ must pass through $O$.

Then $OB$ is always $\frac{1}{2}$, no matter which $A$ we choose on the bottom edge. All that remains is maximizing $AO$. This length is the hypotenuse of a right triangle with the fixed height $MO$, so it is maximized when $AM$ is as large as possible - when $A$ is an endpoint of the bottom edge. Note that $MO = 2 \cdot \frac{\sqrt{3}}{2}$, and that $AM$ can be at most $\frac{1}{2}$, so $AO$ can be at most $\frac{\sqrt{13}}{2}$. So the maximum distance between two points in the diagram is $AO + OB = \frac{1+\sqrt{13}}{2}$.