2. It is easy to see that $h=c-a=\frac{h}{\sin A}-\frac{h}{\sin C}$, simplifying we get
$$
\sin C-\sin A=\sin A \cdot \sin C \text {. }
$$

Given that $A+C=120^{\circ}$, from the above equation we have
$$
2 \sin \frac{C-A}{2} \cos \frac{120^{\circ}}{2}=\frac{1}{2}\left[\cos (C-A)-\cos 120^{\circ}\right] \text {, }
$$

which simplifies to $\left(\sin \frac{C-A}{2}\right)^{2}+\sin \frac{C-A}{2}-\frac{3}{4}=0$,
solving we get $\sin \frac{C-A}{2}=\frac{1}{2}, \sin \frac{C-A}{2}=-\frac{3}{2}$ (discard the latter).
2. In $\triangle A B C$, it is known that the three angles $A, B, C$ form an arithmetic sequence. Suppose the sides opposite to these angles are $a, b, c$, respectively, and $c-a$ equals the altitude $h$ from vertex $A$ to side $AC$, then $\sin \frac{C-A}{2}=$ $\qquad$.