7. In $\triangle A B C$, the sides opposite to $\angle A 、 \angle B 、 \angle C$ are $a 、 b 、 c$ respectively, and $\tan B=\frac{\sqrt{3} a c}{a^{2}+c^{2}-b^{2}}$. Then the size of $\angle B$ is or $\qquad$
II.7. $\frac{\pi}{3}, \frac{2 \pi}{3}$.
By the cosine rule, $b^{2}=a^{2}+c^{2}-2 a c \cos B$. Then $\tan B=\frac{\sqrt{3} a c}{a^{2}+c^{2}-b^{2}}=\frac{\sqrt{3} a c}{2 a c \cos B}=\frac{\sqrt{3}}{2 \cos B}$, i.e., $\sin B=\frac{\sqrt{3}}{2}$.
Therefore, $\angle B$ is $\frac{\pi}{3}$ or $\frac{2 \pi}{3}$.