8. In $\triangle A B C$, the three sides $a, b, c$ form an arithmetic sequence, and $\angle A=3 \angle C, \cos \angle C=$
8. $\frac{1+\sqrt{33}}{8}$.

By the Law of Sines, $a=2 R \sin \angle A, b=2 R \sin \angle B, c=2 R \sin \angle C$, and $\angle A+\angle B+\angle C=180^{\circ}$, then $\angle B=180^{\circ}-4 \angle C>0, \angle C<45^{\circ}$. Given the condition $a+c=2 b$, we have $\sin 3 \angle C+\sin \angle C=2 \sin 4 \angle C$, which simplifies to $2 \sin 2 \angle C \cdot \cos \angle C=4 \sin 2 \angle C \cos 2 \angle C$, so $\cos \angle C=2 \cos 2 \angle C=4 \cos ^{2} \angle C-2$. Therefore, $4 \cos ^{2} \angle C-\cos \angle C-2=0$, solving for $\cos \angle C=\frac{1+\sqrt{33}}{8}, \cos \angle C=\frac{1-\sqrt{33}}{8}$ (discard).