1. (20 points) Given the parabola $y=x^{2}+2 x+m$ intersects the $x$-axis at points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$, where $x_{1}>x_{2}$, and $x_{1}^{2}+x_{2}^{2}=10$.
(1) Find the value of the real number $m$;
(2) Let $M\left(2, y_{0}\right)$ be a point on the parabola $y=x^{2}+2 x+m$. Find a point $P$ on the axis of symmetry of the parabola such that $P A+P M$ is minimized, and determine the coordinates of $P$.
Three, 1. (1) From the conditions, we have
$$
x_{1}+x_{2}=-2, x_{1} x_{2}=m \text {. }
$$

Then $10=x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=4-2 m$
$$
\Rightarrow m=-3 \text {. }
$$

Upon verification, $m=-3$ satisfies the conditions.
(2) From (1), the equation of the parabola is
$$
y=x^{2}+2 x-3 \text {. }
$$

We can obtain that the equation of the axis of symmetry $l$ is $x=-1$, and $A(1,0), B(-3,0)$.
Clearly, the symmetric point of $A$ about the axis of symmetry $l$ is $B$.
Thus, when $P A+P M$ is minimized, $P$ is the intersection point of the line $M B$ and the axis of symmetry $l$.
Also, $y_{0}=2^{2}+2 \times 2-3=5$, i.e., the point $M(2,5)$.
Let the equation of the line $M B$ be $y=k x+b$.
From the conditions, we have
$$
\left\{\begin{array} { l } 
{ 5 = 2 k + b , } \\
{ 0 = - 3 k + b }
\end{array} \Rightarrow \left\{\begin{array}{l}
k=1, \\
b=3
\end{array}\right.\right.
$$
$\Rightarrow$ The equation of the line $M B$ is $y=x+3$.
Let $x=-1$, we get $y=2$, i.e., $P(-1,2)$.
Therefore, $P(-1,2)$ is the point that minimizes the value of $P A+P M$.