7. If $\tan 4 x=$
$$
\frac{\sqrt{3}}{3} \text {, then } \frac{\sin 4 x}{\cos 8 x \cos 4 x}+\frac{\sin 2 x}{\cos 4 x \cos 2 x}+\frac{\sin x}{\cos 2 x \cos x}+\frac{\sin x}{\cos x}=
$$
$\qquad$
$$
\frac{\sin x}{\cos x}+\frac{\sin x}{\cos 2 x \cos x}=\frac{\sin x(\cos 2 x+1)}{\cos 2 x \cos x}=\frac{2 \sin x \cos ^{2} x}{\cos 2 x \cos x}=\frac{\sin 2 x}{\cos 2 x},
$$

Therefore, the original expression $=\frac{\sin 8 x}{\cos 8 x}=\tan 8 x=\frac{2 \tan 4 x}{1-\tan ^{2} 4 x}=\sqrt{3}$.