$\begin{aligned} & \int_0^{\pi / 3} \cos ^4 x d x \\ & =\int_0^{\pi / 3}\left(\frac{1+\cos 2 x}{2}\right)^2 d x\end{aligned}$
$\begin{aligned} & =\frac{1}{4} \int_0^{\pi / 3}\left(1+2 \cos 2 x+\cos ^2 2 x\right) d x \\ & =\frac{1}{4}\left[\int_0^{\pi / 3} \mathrm{dx}+2 \int_0^{\pi / 3} \cos 2 x d x+\int_0^{\pi / 3} \frac{1+\cos 4 x}{2} d x\right] \\ & =\frac{1}{4}\left[\frac{\pi}{3}+(\sin 2 x)_0^{\pi / 3}+\frac{1}{2}\left(\frac{\pi}{3}\right)+\frac{1}{8}(\sin 4 x)_0^{\pi / 3}\right] \\ & =\frac{1}{4}\left[\frac{\pi}{3}+(\sin 2 x)_0^{\pi / 3}+\frac{1}{2}\left(\frac{\pi}{3}\right)+\frac{1}{8}(\sin 4 x)_0^{\pi / 3}\right] \\ & =\frac{1}{4}\left[\frac{\pi}{2}+\frac{\sqrt{3}}{2}+\frac{1}{8} \times\left(-\frac{\sqrt{3}}{2}\right)\right] \\ & =\frac{\pi}{2}+\frac{7 \sqrt{3}}{64} \\ & \therefore \mathrm{a}=\frac{1}{8} ; \mathrm{b}=\frac{7}{64} \\ & \therefore 9 \mathrm{a}+8 \mathrm{~b}=\frac{9}{8}+\frac{7}{8}=2\end{aligned}$