Three, (20 points) The equation $x^{2}+(2 m-1) x+(m-6)=0$ has one root no greater than -1, and the other root no less than 1.
(1) (8 points) Find the range of values for $m$;
(2) (12 points) Find the maximum and minimum values of the sum of the squares of the roots of the equation.
$$
\text { Three, } \begin{aligned}
\because \Delta & =(2 m-1)^{2}-4(m-6) \\
& =4(m-1)^{2}+21>0
\end{aligned}
$$
$\therefore$ The equation always has two distinct real roots.
(1) According to the problem.
Let the quadratic function $y=$ $x^{2}+(2 m-1) x$ $+(m-6)$ have the following properties:
[As shown in the figure, then we have
$$
\left\{\begin{array} { l } 
{ f ( - 1 ) \leqslant 0 , } \\
{ f ( 1 ) \leqslant 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m \geqslant-4, \\
m \leqslant 2
\end{array} \Rightarrow-4 \leqslant m \leqslant 2 .\right.\right.
$$
(2) Let the two roots of the equation be $x_{1}, x_{2}$, then
$$
x_{1}+x_{2}=1-2 m, x_{1} x_{2}=m-6 \text {. }
$$

Thus, $x_{1}^{2}+x_{2}^{2}=4 m^{2}-6 m+13=4\left(m-\frac{3}{4}\right)^{2}+10 \frac{3}{4}$.
When $-4 \leqslant m \leqslant 2$, $f(-4)=101, f(2)=17$.
Therefore, the minimum value of $x_{1}^{2}+x_{2}^{2}$ is $10 \frac{3}{4}$, and the maximum value is 101.