$$
\begin{array}{l}
\text { Three. (20 points) (1) The quadratic function } \\
f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}, a \neq 0)
\end{array}
$$

satisfies
(i) For $x \in \mathbf{R}$, there is $4 x \leqslant f(x) \leqslant \frac{1}{2}(x+2)^{2}$

always holds;
(ii) $f(-4+2 \sqrt{3})=0$.
Find $f(x)$.
(2) Let $f_{1}(x)=\frac{3}{2+x}, f_{n+1}(x)=f_{1}\left(f_{n}(x)\right)$, find $f_{2009}(0)$.
Three, (1) Let $x=2$, we get $8 \leqslant f(2) \leqslant 8$.
Thus, $f(2)=8$,
$4 a+2 b+c=8 \Rightarrow b=\frac{8-4 a-c}{2}$.
Also, $4 x \leqslant f(x)$, so $a x^{2}+(b-4) x+c \geqslant 0$
always holds. Then
$$
\begin{array}{l}
\Delta=(b-4)^{2}-4 a c \\
=\left(\frac{8-4 a-c}{2}-4\right)^{2}-4 a c=\left(2 a-\frac{c}{2}\right)^{2} .
\end{array}
$$
$\leqslant 0$.
Therefore, $c=4 a, b=4-4 a$.
Substituting into $f(-4+2 \sqrt{3})=0$ yields $a=\frac{1}{3}$.
Thus, $b=\frac{8}{3}, c=\frac{4}{3}$.
Hence, $f(x)=\frac{x^{2}+8 x+4}{3}$.
(2) Construct the sequence $a_{n}=\frac{f_{n}(0)-1}{f_{n}(0)+3}$.
Then $\frac{a_{n+1}}{a_{n}}=\frac{f_{n+1}(0)-1}{f_{n+1}(0)+3} \times \frac{f_{n}(0)+3}{f_{n}(0)-1}=-\frac{1}{3}$
(because $\left.f_{n+1}(0)=\frac{3}{2+f_{n}(0)}\right)$. Therefore, the sequence $\left\{a_{n}\right\}$ is a geometric sequence, and $a_{n}=\frac{1}{9} \times\left(-\frac{1}{3}\right)^{n-1}$.
Thus, $a_{2000}=\frac{1}{3^{2010}}$.
From $\frac{f_{200}(0)-1}{f_{2009}(0)+3}=\frac{1}{3^{2010}}$, we solve to get
$$
f_{2000}(0)=\frac{3^{2010}+3}{3^{2010}-1} .
$$