7. In the acute triangle $\triangle A B C$, it is known that $\frac{\sin A}{\sin B}+\cos C=0, \tan A=\frac{\sqrt{2}}{4}$, then $\tan B=$
Solution: $\frac{\sqrt{2}}{2}$.
From $\frac{\sin A}{\sin B}+\cos C=0$, we have
$$
\begin{aligned}
\cos (A+B) \cdot \sin B= & \sin A=\sin ((A+B)-B)=\sin (A+B) \cos B-\cos (A+B) \sin B \\
& \Rightarrow 2 \cos (A+B) \cdot \sin B=\sin (A+B) \cdot \cos B
\end{aligned}
$$
$$
\Rightarrow \tan (A+B)=2 \tan B
$$

Also, $\tan A=\tan ((A+B)-B)=\frac{\tan (A+B)-\tan B}{1+\tan (A+B) \tan B}=\frac{\tan B}{1+2 \tan ^{2} B}=\frac{\sqrt{2}}{4}$
which means $\cot B+2 \tan B=2 \sqrt{2}$, but $\cot B+2 \tan B \geq 2 \sqrt{\cot B \cdot 2 \tan B}=2 \sqrt{2}$, so $\cot B=2 \tan B$. Therefore, $\tan B=\frac{\sqrt{2}}{2}$.