Three. (20 points) Let the domain of the function $f(x)$ be $(0,+\infty)$, and for any positive real numbers $x, y$, the following always holds:
$$
f(x y)=f(x)+f(y)
$$

It is known that $f(2)=1$, and when $x>1$, $f(x)>0$.
(1) Find the value of $f\left(\frac{1}{2}\right)$;
(2) Determine the monotonicity of $y=f(x)$ on $(0,+\infty)$;
(3) A sequence $\left\{a_{n}\right\}$ with all positive terms satisfies $f\left(S_{n}\right)=f\left(a_{n}\right)+f\left(a_{n}+1\right)-1\left(n \in \mathbf{N}_{+}\right)$, where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Find $S_{n}$ and $a_{n}$.
(1) Let $x=y=1$, then $f(1)=f(1)+f(1)$, so $f(1)=0$.
Let $x=2, y=\frac{1}{2}$, then $f(1)=f(2)+f\left(\frac{1}{2}\right)$, so $f\left(\frac{1}{2}\right)=-1$.
(2) Suppose $0 < x_1 < x_2$.
Then $f\left(\frac{x_{1}}{x_{2}}\right)>0$, which means $f\left(x_{1}\right)>f\left(x_{2}\right)$.
Therefore, $y=f(x)$ is monotonically increasing on $(0,+\infty)$.
(3) Since $f(2)=1$, from
$$
f\left(S_{n}\right)+1=f\left(a_{n}\right)+f\left(a_{n}+1\right),
$$

we have $f\left(S_{n}\right)+f(2)=f\left(a_{n}\right)+f\left(a_{n}+1\right)$.
Thus, $f\left(2 S_{n}\right)=f\left(a_{n}\left(a_{n}+1\right)\right)$.
Therefore, $2 S_{n}=a_{n}\left(a_{n}+1\right)$.
So, $S_{n}=\frac{a_{n}^{2}+a_{n}}{2}$.
When $n \geqslant 2$,
$a_{n}=S_{n}-S_{n-1}=\frac{a_{n}^{2}+a_{n}}{2}-\frac{a_{n-1}^{2}+a_{n-1}}{2}$.
Then $\left(a_{n}+a_{n-1}\right)\left(a_{n}-a_{n-1}-1\right)=0$.
By the given condition, $a_{n}-a_{n-1}=1(n \geqslant 2)$.
Also, since $S_{1}=\frac{a_{1}^{2}+a_{1}}{2}, a_{1}=1$, we have $a_{n}=n$.
Therefore, $S_{n}=\frac{n(n+1)}{2}$.