One, (20 points) Find the solution to the following mixed system of equations in $x, y$
$$
\left\{\begin{array}{l}
\frac{x^{2}+3}{2}=x-\cos (x+y), \\
100 \mathrm{e} x \leqslant y \leqslant \frac{100 \pi}{x},
\end{array}\right.
$$

where, $\mathrm{e} \approx 2.71828, \pi \approx 3.14159$.
From the first equation, we have
$$
(x-1)^{2}+2=-2 \cos (x+y) \text {. }
$$

Since $(x-1)^{2}+2 \geqslant 2,-2 \cos (x+y) \leqslant 2$,
it follows that $x=1$, and $\cos (x+y)=-1$.
From $\cos (1+y)=-1$, we get
$1+y=(2 k+1) \pi(k \in \mathbf{Z})$.
Thus, $x=1, y=(2 k+1) \pi-1(k \in \mathbf{Z})$.
Substituting (1) into the second inequality, we get
$$
100 e \leqslant(2 k+1) \pi-1 \leqslant 100 \pi \text {, }
$$

which simplifies to $\frac{100 \mathrm{e}-\pi+1}{2 \pi} \leqslant k \leqslant \frac{99 \pi+1}{2 \pi}$.
Solving this, we get $42.92 \leqslant k \leqslant 49.66$.
Thus, $k=43,44, \cdots, 49$.
Therefore, the solution for the mixed group is
$$
\left\{\begin{array}{l}
x=1, \\
y=(2 k+1) \pi-1(k=43,44, \cdots, 49) .
\end{array}\right.
$$