Four. (20 points) Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms as $S_{n}$, and it satisfies $a_{2}=2, S_{n}=\frac{n\left(1+a_{n}\right)}{2}\left(n \in \mathbf{N}_{+}\right)$.
(1) Find the general term of the sequence $\left\{a_{n}\right\}$.
(2) If $b_{n}=a_{n}^{\frac{i}{a_{n}+1}}$, find the maximum term of the sequence $\left\{b_{n}\right\}$.
(3) For the sequence $\left\{b_{n}\right\}$ in (2), does there exist $b_{n}=$ $b_{m}(n \neq m)$? If so, find all equal pairs; if not, explain the reason.
(1) From $S_{1}=\frac{1+a_{1}}{2}=a_{1}$, we know $a_{1}=1$.
When $n>1$,
$$
a_{n}=S_{n}-S_{n-1}=\frac{n\left(1+a_{n}\right)}{2}-\frac{(n-1)\left(1+a_{n-1}\right)}{2} \text {. }
$$

Simplifying, we get $(n-2) a_{n}-(n-1) a_{n-1}+1=0$.
Substituting $n+1$ for $n$ gives
$$
(n-1) a_{n+1}-n a_{n}+1=0 \text {. }
$$

Subtracting the two equations, we get
$$
(n-1) a_{n+1}-2(n-1) a_{n}+(n-1) a_{n-1}=0 \text {. }
$$

Thus, $a_{n+1}-2 a_{n}+a_{n-1}=0$.
Therefore, $\left\{a_{n}\right\}$ is an arithmetic sequence.
Given $a_{1}=1, a_{2}=2$, we have $a_{n}=n$.
(2) $b_{n}=n^{\frac{1}{n+1}}$, consider the range of $n$ when $b_{n}>b_{n+1}$.
Notice that $n^{\frac{1}{n+1}}>(n+1)^{\frac{1}{n+2}}$, i.e.,
$$
n^{n+2}>(n+1)^{n+1} \text {. }
$$

Then $n>\left(\frac{n+1}{n}\right)^{n+1}=\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n}\right)^{n}$.
When $n \geqslant 3$,
$$
\begin{array}{l}
\left(1+\frac{1}{n}\right)^{n} \\
=1+\mathrm{C}_{n}^{1} \cdot \frac{1}{n}+\mathrm{C}_{n}^{2} \cdot \frac{1}{n^{2}}+\cdots+\mathrm{C}_{n}^{k} \cdot \frac{1}{n^{k}}+\cdots+\mathrm{C}_{n}^{n} \cdot \frac{1}{n^{n}} \\
=1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\cdots+ \\
\quad \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \cdots\left(1-\frac{n-k+1}{n}\right)+ \\
\quad \cdots+\frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \cdots\left(1-\frac{n-1}{n}\right) \\
3\left(1+\frac{1}{n}\right)=3+\frac{3}{n}>3$, i.e., when $n \geqslant 4$, we have $b_{n}>b_{n+1}$.
By comparing the sizes of $b_{1}, b_{2}, b_{3}, b_{4}, b_{5}$, we know $b_{1}b_{5}$.
Therefore, the sequence $\left\{b_{n}\right\}$ satisfies
$$
b_{1}b_{5}>b_{6}>\cdots \text {. }
$$

Thus, the maximum term of the sequence $\left\{b_{n}\right\}$ is $b_{4}$.
(3) Clearly, $b_{1}=1$.

From $b_{n}=n^{\frac{1}{n+1}}$, we know that when $n \neq 1$, $b_{n} \neq 1$.
From equation (1), if the sequence $\left\{b_{n}\right\}$ has two equal terms, they can only be $b_{2}, b_{3}$ and subsequent terms.

Also, $b_{2}=2^{\frac{1}{3}}=8^{\frac{1}{9}}=b_{8}$, i.e., the 2nd term is equal to the 8th term.
From equation (1), only the 8th term is equal to the 2nd term.
And $b_{3}=3^{\frac{1}{4}}>b_{5}=5^{\frac{1}{6}}$, so from equation (1), there is no term equal to the 3rd term.

Therefore, the sequence $\left\{b_{n}\right\}$ has only one pair of equal terms $b_{2}=b_{8}$.