11. (20 points) The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=\sqrt{a_{n}^{2}-2 a_{n}+3}+c\left(n \in \mathbf{N}^{*}\right)$. Here, $c$ is a constant greater than 0.
(1) If $c=1$, find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If the sequence $\left\{a_{n}\right\}$ has an upper bound, find the range of values for $c$.
11. (1) It is easy to see that $a_{n+1}=\sqrt{a_{n}^{2}-2 a_{n}+3}+1 \geqslant 1$. Therefore, $a_{n} \geqslant 1\left(n \in \mathbf{N}^{*}\right)$, and $\left(a_{n+1}-1\right)^{2}=\left(a_{n}-1\right)^{2}+2$.
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. Hence, $b_{n}=2(n-1)$.
Since $a_{n} \geqslant 1$, we have $a_{n}=1+\sqrt{2(n-1)}$.
(2) When $c \geqslant 1$, we know $a_{1}=1, a_{n+1} \geqslant \sqrt{a_{n}^{2}-2 a_{n}+3}+1\left(n \in \mathbf{N}^{*}\right)$. Therefore, $a_{n} \geqslant 1\left(n \in \mathbf{N}^{*}\right)$, and $\left(a_{n+1}-1\right)^{2} \geqslant\left(a_{n}-1\right)^{2}+2$. Then $\left(a_{n}-1\right)^{2} \geqslant 2(n-1) \Rightarrow a_{n} \geqslant 1+\sqrt{2(n-1)}$, at this time the sequence is unbounded.

When $0 < c < 1$, we have $a_{1}=1, a_{n+1}=\sqrt{a_{n}^{2}-2 a_{n}+3}+c \geqslant \sqrt{\left(a_{n}-1\right)^{2}+2} \geqslant \sqrt{2}$, hence $a_{n} \geqslant 1\left(n \in \mathbf{N}^{*}\right)$.
We will use mathematical induction to prove: $a_{n}<\frac{1+c}{2}+\frac{1}{1-c}$,
Thus, when $n=1$, the conclusion holds.
Assume that when $n=k\left(k \in \mathbf{N}^{*}\right)$, the conclusion holds, i.e., $a_{k}<\frac{1+c}{2}+\frac{1}{1-c}$. Then when $n=k+1$,
\[
\begin{aligned}
a_{k+1} & =\sqrt{\left(a_{k}-1\right)^{2}+2}+c<\sqrt{\left(\frac{1+c}{2}+\frac{1}{1-c}-1\right)^{2}+2}+c \\
& =\frac{1}{1-c}+\frac{1-c}{2}+c=\frac{1+c}{2}+\frac{1}{1-c}
\end{aligned}
\]
Thus, when $n=k+1$, the conclusion holds.
In summary, the range of values for $c$ is $(0,1)$.