One. (20 points) Given the one-variable quartic equation $x^{4}+4 x^{3}+(k+7) x^{2}+2(k+3) x+3 k=0$ has real roots.
(1) Find the range of values for $k$;
(2) If the product of all real roots of the equation is -4, and $m, n$ are two of the real roots, find the value of $-m^{3}-\frac{32}{n^{2}}-4 n+\frac{4}{n}$.
(1) A quartic equation can be transformed into
$$
\begin{array}{l}
x^{2}\left(x^{2}+2 x+3\right)+2 x\left(x^{2}+2 x+3\right)+ \\
k\left(x^{2}+2 x+3\right)=0,
\end{array}
$$

which is $\left(x^{2}+2 x+3\right)\left(x^{2}+2 x+k\right)=0$.
Therefore, the quartic equation can be reduced to two quadratic equations:
$$
x^{2}+2 x+3=0
$$

or $x^{2}+2 x+k=0$.
Given that the quartic equation has real roots and equation (1) has no real roots, it follows that equation (2) must have real roots. Thus,
$$
\Delta=2^{2}-4 k \geqslant 0 \text {. }
$$

Therefore, $k \leqslant 1$.
(2) From (1), the product of all real roots of the quartic equation is the product of all real roots of equation (2).
By Vieta's formulas, we get $k=-4$.
Thus, $m$ and $n$ are the two real roots of the quadratic equation $x^{2}+2 x-4=0$.
Hence, $\left\{\begin{array}{l}m^{2}+2 m-4=0, \\ m+n=-2, \\ m n=-4, \\ m=\frac{-2 \pm \sqrt{2^{2}-4 \times(-4)}}{2}=-1 \pm \sqrt{5} .\end{array}\right.$
Then $-m^{3}-\frac{32}{n^{2}}-4 n+\frac{4}{n}$
$$
\begin{array}{l}
=-m^{3}-2 m^{2}-4 n+(-m) \\
=-m\left(m^{2}+2 m-4\right)-4 m-4 n-m \\
=-m \times 0-4 \times(-2)-m \\
=8-m=8-(-1 \pm \sqrt{5})=9 \mp \sqrt{5} .
\end{array}
$$