Three. (20 points) Given a function $f(x)$ defined on $\mathbf{R}$ that satisfies $f\left(f(x)-x^{2}+x\right)=f(x)-x^{2}+x$.
(1) When $f(2)=3$, find $f(1)$; when $f(0)=a$, find $f(a)$.
(2) If there is exactly one real number $x_{0}$ such that $f\left(x_{0}\right)=x_{0}$, find the analytical expression of the function $f(x)$.
(1) Let $x=2$, we get
$$
f\left(f(2)-2^{2}+2\right)=f(2)-2^{2}+2 \text {. }
$$

Since $f(2)=3$, then
$$
f\left(3-2^{2}+2\right)=3-2^{2}+2 \Rightarrow f(1)=1 \text {. }
$$

Let $x=0$, we get $f(f(0))=f(0)$.
Since $f(0)=a$, then $f(a)=a$.
(2) For any $x \in \mathbf{R}$, we have
$$
f\left(f(x)-x^{2}+x\right)=f(x)-x^{2}+x \text {, }
$$

and there is exactly one real number $x_{0}$ such that $f\left(x_{0}\right)=x_{0}$, so,
$$
f(x)-x^{2}+x=x_{0} \text {. }
$$

Let $x=x_{0}$, we get $f\left(x_{0}\right)-x_{0}^{2}+x_{0}=x_{0}$.
Since $f\left(x_{0}\right)=x_{0}$, then $x_{0}-x_{0}^{2}=0$. Therefore,
$x_{0}=0$ or $x_{0}=1$.
If $x_{0}=0$, then $f(x)-x^{2}+x=0$, i.e., $f(x)=x^{2}-x$. But $x^{2}-x=x$ has two distinct real roots, which is a contradiction.

If $x_{0}=1$, then $f(x)-x^{2}+x=1$, i.e., $f(x)=x^{2}-x+1$. Clearly, the equation $x^{2}-x+1=x$ has only one real root, which meets the requirement. In summary, the function we seek is $f(x)=x^{2}-x+1(x \in \mathbf{R})$.