1. Given the real-coefficient equations about $x$: $x^{2}-2 x+2=0$ and $x^{2}+2 m x+1=0$, the four distinct roots of these equations correspond to points in the complex plane that lie on a circle. Then the set of values for $m$ is
II. 1. $|m|-10$, i.e., $m1$ when the equation $x^{2}+2 m x+1=0$ has two distinct real roots $x_{3}, x_{4}$, then the points corresponding to $x_{1}, x_{2}$ lie on the circle with the diameter endpoints corresponding to $x_{3}, x_{4}$. By the equation of the circle with diameter, the equation of the circle is
$$
\left(x-x_{3}\right)\left(x-x_{4}\right)+y^{2}=0 \text {, }
$$

i.e., $x^{2}+y^{2}-\left(x_{3}+x_{4}\right) x+x_{3} x_{4}=0$.
Substituting $x_{3}+x_{4}=-2 m, x_{3} x_{4}=1$ and the coordinates of the points corresponding to $x_{1}, x_{2}$, $(1, \pm 1)$, into the equation, we get $m=-\frac{3}{2}$.
Therefore, the set of values for $m$ is
$$
\left\{m \mid-1<m<1 \text { or } m=-\frac{3}{2}\right\} \text {. }
$$