Answer: $\frac{7 \sqrt{3}}{4}$
Solution: As shown in the figure, rotate $\triangle A B O$ 60 degrees clockwise around point $B$.
It is easy to prove that $\triangle B O D$ is an equilateral triangle, so $O D=O B=3, C D=O A=4$.
Since
$$
3^{2}+4^{2}=5^{2},
$$

$\triangle \mathrm{COD}$ is a right triangle,

we can get
$$
S_{\triangle A O B}+S_{\triangle B O C}=S_{B O C D}=S_{\triangle B O D}+S_{\triangle C O D}=\frac{9 \sqrt{3}}{4}+6 。
$$

Similarly, rotate $\triangle A B O$ 60 degrees counterclockwise around point $A$, we can get
$$
S_{\triangle A O B}+S_{\triangle A O C}=S_{A O C E}=S_{\triangle A O E}+S_{\triangle C O E}=4 \sqrt{3}+6 \text {, }
$$

thus $S_{\triangle A O C}-S_{\triangle B O C}=\left(S_{\triangle A O B}+S_{\triangle A O C}\right)-\left(S_{\triangle A O B}+S_{\triangle B O C}\right)=\frac{7 \sqrt{3}}{4}$ .
66. Point $O$ is inside equilateral $\triangle A B C$, if $O B=3, O A=4, O C=5$, then the difference in area between $\triangle A O C$ and $\triangle B O C$ is