1. (16 points) Given the equation in $x$
$$
x^{2}-\left(m^{2}+2 m-3\right) x+2(m+1)=0
$$

has two real roots that are opposites of each other.
(1) Find the value of the real number $m$;
(2) If the equation in $x$ $x^{2}-(k+m) x-3 m-k-5=0$ has roots that are all integers, find all real numbers $k$ that satisfy the condition.
Three, 1. (1) From the problem, we know that $m^{2}+2 m-3=0$. Solving this, we get $m=-3, m=1$.
When $m=-3$, $\Delta=16>0$; when $m=1$, $\Delta=-16<0$. Therefore, the value of the real number $m$ is -3.
(2) From (1), $m=-3$, then the equation becomes $x^{2}-(k-3) x+4-k=0$. Let the two roots of the equation be $x_{1}$ and $x_{2}$, so we have $x_{1}+x_{2}=k-3, x_{1} x_{2}=4-k$. Then $x_{1} x_{2}+x_{1}+x_{2}=1$, which means $\left(x_{1}+1\right)\left(x_{2}+1\right)=2$.
Since $x_{1}$ and $x_{2}$ are integers, we have $\left\{\begin{array}{l}x_{1}+1=1, \\ x_{2}+1=2\end{array}\right.$ or $\left\{\begin{array}{l}x_{1}+1=-1, \\ x_{2}+1=-2 .\end{array}\right.$ Solving these, we get $\left\{\begin{array}{l}x_{1}=0, \\ x_{2}=1\end{array}\right.$ or $\left\{\begin{array}{l}x_{1}=-2, \\ x_{2}=-3 .\end{array}\right.$ Therefore, $k=4$ or $k=-2$.