8. Let $\triangle A B C$ have three interior angles in arithmetic progression, and the reciprocals of the three corresponding sides $a, b, c$ also form an arithmetic progression. Try to find $A, B, C$.
In $\triangle A B C$, the three interior angles $A, B, C$ form an arithmetic sequence $\Rightarrow B=\frac{\pi}{3}$,
and $\frac{1}{a}+\frac{1}{c}=\frac{2}{b} \Rightarrow \frac{1}{\sin A}+\frac{1}{\sin C}=\frac{2}{\sin B}=\frac{4}{\sqrt{3}} \Rightarrow \frac{\sin A+\sin C}{\sin A \sin C}=\frac{4}{\sqrt{3}}$
$\Rightarrow \frac{2 \sin \frac{A+C}{2} \cos \frac{A-C}{2}}{\frac{1}{2}[\cos (A-C)-\cos (A+C)]}=\frac{4 \sqrt{3} \cos \frac{A-C}{2}}{2 \cos (A-C)+1}=\frac{4 \sqrt{3}}{3}$
$\Rightarrow 4 \cos ^{2} \frac{A-C}{2}-3 \cos \frac{A-C}{2}-1=0 \Rightarrow \cos \frac{A-C}{2}=1$ or $-\frac{1}{4}$.
When $\cos \frac{A-C}{2}=-\frac{1}{4}$, $\left|\frac{A-C}{2}\right|>\frac{\pi}{2} \Rightarrow|A-C|>\pi$, contradiction;
Therefore, $\cos \frac{A-C}{2}=1 \Rightarrow A=C \Rightarrow A=B=C=\frac{\pi}{3}$.