1. (20 points) Given the parabola $y=x^{2}$ and the line $y=(k+2) x-(2 k-1)$.
(1) Prove: regardless of what real number $k$ is, the parabola and the line always have two different intersection points;
(2) Let the two different intersection points of the parabola and the line be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, if $x_{1} 、 x_{2}$ are both integers, find the value of the real number $k$.
$$
\begin{array}{l}
\text { Three, 1. (1) From } \\
\left\{\begin{array}{l}
y=x^{2}, \\
y=(k+2) x-(2 k-1),
\end{array}\right.
\end{array}
$$

eliminating $y$ yields
$$
\begin{array}{l}
x^{2}-(k+2) x+(2 k-1)=0 . \\
\text { Since } \Delta=(k+2)^{2}-4(2 k-1) \\
=k^{2}-4 k+8=(k-2)^{2}+4>0,
\end{array}
$$

therefore, the parabola and the line always have two distinct intersection points.
(2) From (1) and the relationship between roots and coefficients, we get
$$
x_{1}+x_{2}=k+2, x_{1} x_{2}=2 k-1 \text {. }
$$

eliminating $k$ yields
$$
\begin{array}{l}
x_{1} x_{2}-2 x_{1}-2 x_{2}=-5 \\
\Rightarrow\left(x_{1}-2\right)\left(x_{2}-2\right)=-1 .
\end{array}
$$

Assume $x_{1}<x_{2}$. Then
$$
\left\{\begin{array} { l } 
{ x _ { 1 } - 2 = - 1 , } \\
{ x _ { 2 } - 2 = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x_{1}=1, \\
x_{2}=3 .
\end{array}\right.\right.
$$

Thus, $k=2$.