One. (20 points) Given the parabola $y=x^{2}+3 x+c$ passes through two points $(m, 0)$ and $(n, 0)$, and
$$
m^{3}+3 m^{2}+(c-2) m-2 n-c=8,
$$

the parabola intersects the hyperbola $y=\frac{k}{x}$ at the point $(1, d)$.
(1) Find the equations of the parabola and the hyperbola;
(2) Given points $P_{1}, P_{2}, \cdots, P_{2008}$ are all on the hyperbola $y=\frac{k}{x}(x>0)$, with their x-coordinates being $a$, $2a, \cdots, 2008a$, and $O$ being the origin, let $S_{1}=$ $S_{\triangle P_{1} P_{2} O}, S_{2}=S_{\triangle P_{1} P_{3} O}, \cdots \cdots$ Point $Q$ is on the hyperbola $y=\frac{k}{x}(x<0)$, and a perpendicular line $Q M \perp y$-axis is drawn from $Q$ to $M$, let $S=S_{\triangle Q M O}$. Find
$$
S_{1}+S_{2}+\cdots+S_{2007}+\frac{S}{2}+\frac{S}{3}+\cdots+\frac{S}{2008}
$$

the value.
(1) According to the problem, we have
$$
\left\{\begin{array}{l}
m\left(m^{2}+3 m+c\right)-2(m+n)-c=8 \\
m^{2}+3 m+c=0, \\
m+n=-3 .
\end{array}\right.
$$

Substituting equations (2) and (3) into equation (1) gives $c=-2$.
Therefore, the equation of the parabola is $y=x^{2}+3 x-2$.
Since the intersection point of the parabola and the hyperbola is $(1, d)$, we have
$$
\left\{\begin{array}{l}
d = 1^{2} + 3 \times 1 - 2, \\
d = \frac{k}{1}
\end{array} \Rightarrow \left\{\begin{array}{l}
d=2, \\
k=2 .
\end{array}\right.\right.
$$

Therefore, the equation of the hyperbola is $y=\frac{2}{x}$.
(2) Since points $P_{1}$ and $P_{n+1} (n=1,2, \cdots, 2007)$ are on the hyperbola $y=\frac{2}{x} (x>0)$, and their x-coordinates are $a$ and $(n+1)a$ respectively, the y-coordinates of points $P_{1}$ and $P_{n+1}$ are $\frac{2}{a}$ and $\frac{2}{(n+1)a}$ respectively.

As shown in Figure 6, draw a line $a_{1} \parallel x$-axis through point $P_{1}$ intersecting the y-axis at $A_{1}$, and draw a line $b_{n+1} \parallel y$-axis through $P_{n+1}$ intersecting the x-axis at $B_{n+1}$ and intersecting $a_{1}$ at $C_{n+1}$. Then
$$
\begin{aligned}
& S_{n}=S_{P_{1} P_{n+1} o} \\
= & (n+1) a \cdot \frac{2}{a}-\frac{1}{2} a \cdot \frac{2}{a}- \\
& \frac{1}{2}(n+1) a \cdot \frac{2}{(n+1) a}- \\
& \frac{1}{2}[(n+1) a-a]\left[\frac{2}{a}-\frac{2}{(n+1) a}\right] \\
= & n+\frac{n}{n+1} .
\end{aligned}
$$

Let $Q(e, f) (e<0)$. Then $f=\frac{2}{e}$.
Thus, $S=S_{\triangle Q M O}=\frac{1}{2}|e| \cdot|f|=\frac{1}{2} e f=1$.
$$
\begin{array}{l}
\text { Therefore, } S_{1}+S_{2}+\cdots+S_{2007}+\frac{S}{2}+\frac{S}{3}+\cdots+\frac{S}{2008} \\
=\left(1+\frac{1}{2}\right)+\left(2+\frac{2}{3}\right)+\cdots+ \\
\left(2007+\frac{2007}{2008}\right)+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2008} \\
= 1+2+\cdots+2007+1 \times 2007 \\
= \frac{(1+2007) \times 2007}{2}+2007 \\
= 2017035 .
\end{array}
$$