Three. (20 points) Given the function
$$
f(x)=-x^{3}+a x^{2}+b x+c \quad (a, b, c \in \mathbf{R})
$$

is monotonically decreasing on the interval $(-\infty, 0)$, monotonically increasing on the interval $(0,1)$, and $f(x)$ has three zeros on $\mathbf{R}$, with 1 being one of them.
(1) Find the range of $f(2)$;
(2) Discuss the number of common points between the line $l: y=x-1$ and the curve $C$: $y=f(x)$.
(1) From the given conditions, we have
$$
f^{\prime}(x)=-3 x^{2}+2 a x+b \text {. }
$$

Since $f(x)$ is decreasing on the interval $(-\infty, 0)$ and increasing on the interval $(0,1)$, it follows that $f^{\prime}(0)=0$, which gives $b=0$.
Also, since $f(1)=0$, we have $c=1-a$.
Thus, $f(x)=-x^{3}+a x^{2}+1-a$,
and $f^{\prime}(x)=-3 x^{2}+2 a x$.
Setting $f^{\prime}(x)=0$, we get $x_{1}=0, x_{2}=\frac{2 a}{3}$.
Since $f(x)$ is increasing on the interval $(0,1)$ and $f(x)$ has three zeros, it follows that $x_{2}=\frac{2 a}{3}>1$, i.e., $a>\frac{3}{2}$.

Therefore, $f(2)=3 a-7>-\frac{5}{2}$, which means the range of $f(2)$ is $\left(-\frac{5}{2},+\infty\right)$.
(2) From the system of equations $\left\{\begin{array}{l}y=x-1, \\ y=-x^{3}+a x^{2}+1-a,\end{array}\right.$, we get $(x-1)\left(x^{2}+(1-a) x+2-a\right)=0$
$\Rightarrow x=1$ or
$$
x^{2}+(1-a) x+2-a=0\left(a>\frac{3}{2}\right) \text {. }
$$

The discriminant of equation (1) is $\Delta=a^{2}+2 a-7\left(a>\frac{3}{2}\right)$.
When $\Delta>0$, i.e., $a>2 \sqrt{2}-1$, equation (1) has two distinct real roots.
When $a=2$, the two roots of equation (1) are $0$ and $1$.
Therefore, when $\frac{3}{2}<a<2 \sqrt{2}-1$ or $a>2 \sqrt{2}-1$ and $a \neq 2$, the line $l$ intersects the curve $C$ at three points.