II. (50 points) Given an odd function $f(x)$ defined on $(-\infty, 0) \cup (0, +\infty)$, which is decreasing on the interval $(0, +\infty)$, and $f(k)=0$ for some $k>0$. Another function is defined as
$$
\varphi(\alpha)=\sin ^{2} \alpha+(\\xi+1) \cos \alpha-\xi^{2}-\xi-k,
$$
where $\alpha \in [0, \pi]$.
If the set $A=\{\xi \mid \varphi(\alpha) > 0, \forall \alpha \in [0, \pi]\}$,

try to find $A \cap B$.


Given $f(k)=0 \quad(k>0), f(x)$ is an odd function, so $f(-k)=-f(k)=0$.
Also, $f(x)$ is a decreasing function in $(0,+\infty)$. Therefore, $f(x)$ is also a decreasing function in $(-\infty, 0)$.
Thus, the solution set for $f(x)>0$ is
$\{x \mid |x| < 0\}$.
Then $B=\{\xi \mid f(\varphi(\alpha))>0, \forall \alpha \in[0, \pi]\}$
$=\{\xi \mid 0 < 0$.
Since $\alpha \in[0, \pi]$, we have $\cos \alpha \in[-1,1]$.
Let $u=\cos \alpha$, then $u \in[-1,1]$.
Thus, the problem is converted to finding the values of $\xi$ such that the inequality $u^{2}-(\xi+1) u+\xi^{2}+\xi-1>0$ holds for all $u \in[-1,1]$.
Let $g(u)=u^{2}-(\xi+1) u+\xi^{2}+\xi-1$.
(1) When $\Delta<0$, $g(u)>0$ always holds. At this time,
$$
(\xi+1)^{2}-4\left(\xi^{2}+\xi-1\right)=-3 \xi^{2}-2 \xi+5<0
$$
Solving this, we get $\xi>1$ or $\xi<-\frac{5}{3}$.
(2) For $g(u)>0$ to hold for all $u \in[-1,1]$, it can also be determined by
$$
\left\{\begin{array}{l}
\Delta \geqslant 0, \\
\frac{\xi+1}{2} < -1, \\
g(-1)>0
\end{array} \text { or } \left\{\begin{array}{l}
\Delta \geqslant 0, \\
\frac{\xi+1}{2}>1, \\
g(1)>0
\end{array}\right.\right.
$$
Solving this, we get $\xi \in \varnothing$.
Therefore, from (1) and (2), the range of $\xi$ is
$$
\left\{\xi \mid \xi>1 \text { or } \xi<-\frac{5}{3}\right\}.
$$