The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, \\
a_{n+1}=\sqrt{a_{n}^{2}-2 a_{n}+3}+c\left(n \in \mathbf{N}_{+}\right),
\end{array}
$$

where $c$ is a constant greater than 0.
(1) If $c=1$, find the general term of the sequence $\left\{a_{n}\right\}$;
(2) If the sequence has an upper bound, find the range of values for $c$.
(1) It is easy to know that
$$
a_{n+1}=\sqrt{a_{n}^{2}-2 a_{n}+3}+1 \geqslant 1 \text {. }
$$

Therefore, $a_{n} \geqslant 1\left(n \in \mathbf{N}_{+}\right)$.
Then $\left(a_{n+1}-1\right)^{2}=\left(a_{n}-1\right)^{2}+2$.
Let $b_{n}=\left(a_{n}-1\right)^{2}$. Then $\left\{b_{n}\right\}$ is an arithmetic sequence with the first term 0 and common difference 2.
Thus, $b_{n}=2(n-1)$.
Since $a_{n} \geqslant 1$, we have $a_{n}=1+\sqrt{2(n-1)}$.
(2) When $c \geqslant 1$, we know that
$$
\begin{array}{l}
a_{1}=1, \\
a_{n+1} \geqslant \sqrt{a_{n}^{2}-2 a_{n}+3}+1\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$

Therefore, $a_{n} \geqslant 1\left(n \in \mathbf{N}_{+}\right)$, and
$$
\begin{array}{l}
\left(a_{n+1}-1\right)^{2} \geqslant\left(a_{n}-1\right)^{2}+2 . \\
\text { Then }\left(a_{n}-1\right)^{2} \geqslant 2(n-1) \\
\Rightarrow a_{n} \geqslant 1+\sqrt{2(n-1)},
\end{array}
$$

In this case, the sequence is unbounded.
When $0 < c < 1$, we have
$$
\begin{array}{l}
a_{1}=1, \\
a_{n+1}=\sqrt{a_{n}^{2}-2 a_{n}+3}+c \\
\geqslant \sqrt{\left(a_{n}-1\right)^{2}+2} \geqslant \sqrt{2},
\end{array}
$$

Therefore, $a_{n} \geqslant 1\left(n \in \mathbf{N}_{+}\right)$.
We will prove by mathematical induction that:
$$
a_{n}<\frac{1+c}{2}+\frac{1}{1-c} \\
\geqslant 2 \sqrt{\frac{1-c}{2} \cdot \frac{1}{1-c}}=\sqrt{2}>a_{1},
$$

Thus, when $n=1$, the conclusion holds.
Assume that when $n=k\left(k \in \mathbf{N}_{+}\right)$, the conclusion holds, i.e.,
$$
a_{k}<\frac{1+c}{2}+\frac{1}{1-c} \text {. }
$$

Then when $n=k+1$,
$$
\begin{array}{l}
a_{k+1}=\sqrt{\left(a_{k}-1\right)^{2}+2}+c \\
<\sqrt{\left(\frac{1+c}{2}+\frac{1}{1-c}-1\right)^{2}+2}+c \\
=\frac{1}{1-c}+\frac{1-c}{2}+c \\
=\frac{1+c}{2}+\frac{1}{1-c} .
\end{array}
$$

Thus, when $n=k+1$, the conclusion holds.
In summary, the range of $c$ is $(0,1)$.
(Zhang Lei, Northeast Yucai High School, Shenyang City, Liaoning Province, 110179)