10. Let $a_{1}=1, a_{n+1}=\sqrt{a_{n}^{2}-2 a_{n}+2}+b,\left(n \in N^{*}\right)$. (1) If $b=1$, find $a_{2}, a_{3}$ and the general term formula of the sequence $\left\{a_{n}\right\}$; (2) If $b=-1$, does there exist a real number $c$ such that $a_{2 n}<c<a_{2 n+1}$, for all $n \in N^{*}$?
10. Analysis: (1) Solution one: When $b=1$, $\mathrm{Q} a_{n+1}=\sqrt{a_{n}^{2}-2 a_{n}+2}+1, \therefore\left(a_{n+1}-1\right)^{2}=\left(a_{n}-1\right)^{2}+1$
$\therefore\left\{\left(a_{n}-1\right)^{2}\right\}$ is an arithmetic sequence with a common difference of 1 and the first term $\left(a_{1}-1\right)^{2}=0$, hence $\left(a_{n}-1\right)^{2}=n-1$, i.e., $a_{n}=\sqrt{n-1}+1\left(n \in N^{*}\right)$

Solution two: $a_{2}=2, a_{3}=\sqrt{2}+1 . \therefore a_{1}=\sqrt{1-1}+1, a_{2}=\sqrt{2-1}+1, a_{3}=\sqrt{3-1}+1$.
Therefore, we conjecture $a_{n}=\sqrt{n-1}+1$. We will prove this by mathematical induction: When $n=1$, the conclusion is obviously true.
Assume the conclusion holds for $n=k$, i.e., $a_{k}=\sqrt{k-1}+1$. Then $a_{k+1}=\sqrt{\left(a_{k}-1\right)^{2}+1}+1=\sqrt{(k-1)+1}+1=\sqrt{(k+1)-1}+1$.

Thus, the conclusion holds for $n=k+1$. Therefore, $a_{n}=\sqrt{n-1}+1, \quad\left(n \in N^{*}\right)$
(2) Let $f(x)=\sqrt{(x-1)^{2}+1}-1$, then $a_{n+1}=f\left(a_{n}\right)$. Let $c=f(c)$, i.e., $c=\sqrt{(c-1)^{2}+1}-1$, solving gives $c=\frac{1}{4}$.

We will use mathematical induction to prove the strengthened proposition: $a_{2 n}f\left(a_{2 n+1}\right)>f(1)=a_{2}$, i.e., $1>c>a_{2 k+2}>a_{2}$

Since $f(x)$ is a decreasing function on $(-\infty, 1]$, we have $c=f(c)<f\left(a_{2 k+2}\right)<f\left(a_{2}\right)=a_{3}<1$. In summary, $c=\frac{1}{4}$ meets the condition.