Three. (20 points) Given $\sin (2 \alpha+\beta)=3 \sin \beta$. Let $\tan \alpha=x, \tan \beta=y$, and denote $y=f(x)$.
(1) Find the expression for $f(x)$;
(2) Define the sequence of positive numbers $\left\{a_{n}\right\}: a_{1}=\frac{1}{2}, a_{n+1}^{2}=$ $2 a_{n} f\left(a_{n}\right)\left(n \in \mathbf{N}_{+}\right)$. Find the general term formula for the sequence $\left\{a_{n}\right\}$.
(1) From $\sin (2 \alpha+\beta)=3 \sin \beta$, we get $\sin [(\alpha+\beta)+\alpha]=3 \sin [(\alpha+\beta)-\alpha]$,
which means $\sin (\alpha+\beta) \cdot \cos \alpha=2 \cos (\alpha+\beta) \cdot \sin \alpha$.
Therefore, $\tan (\alpha+\beta)=2 \tan \alpha$.
Thus, $\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=2 \tan \alpha$, which simplifies to $\frac{x+y}{1-x y}=2 x$.
Solving for $y$ gives $y=\frac{x}{1+2 x^{2}}$.
Hence, $f(x)=\frac{x}{1+2 x^{2}}$.
(2) Since $a_{n+1}^{2}=2 a_{n} f\left(a_{n}\right)=\frac{2 a_{n}^{2}}{1+2 a_{n}^{2}}$, then
$$
\frac{1}{a_{n+1}^{2}}=\frac{1}{2 a_{n}^{2}}+1 \text {, }
$$

which means $\frac{1}{a_{n+1}^{2}}-2=\frac{1}{2}\left(\frac{1}{a_{n}^{2}}-2\right)$.
Therefore, $\left\{\frac{1}{a_{n}^{2}}-2\right\}$ is a geometric sequence with the first term 2 and common ratio $\frac{1}{2}$. So, $\frac{1}{a_{n}^{2}}-2=2\left(\frac{1}{2}\right)^{n-1}$.
Thus, $a_{n}=\sqrt{\frac{2^{n-2}}{2^{n-1}+1}}$.