Problem 245

Solution:
(I) Given $f(x) = \begin{cases} x+ \frac {1}{x}-1, & (x>0) \\ -x^{3}+1, & (x\leq0)\end{cases}$,
For $x>0$, according to the graph and properties of the root function,
we have: when $x=1$, the function reaches its minimum value of 1,
For $x\leq0$, the function is decreasing,
when $x=0$, the function reaches its minimum value of 1,
Overall, the minimum value of the function $f(x)$ is $\boxed{1}$.
(II) From (I), we have $m^2+2m-2\leq1$ holds for any $x\in\mathbb{R}$
which means $m^2+2m-3\leq0$, solving this gives $-3\leq m\leq1$,
Therefore, for proposition $p$: $-3\leq m\leq1$.
For proposition $q$: The exponential function $y=(m^2-1)^x$ is increasing,
thus $m^2-1>1$,
Therefore, for proposition $q$: $m\sqrt{2}$.
If "p or q" is true, and "p and q" is false,
then $p$, $q$ are true and false respectively,
Considering two cases:
If $p$ is true and $q$ is false, then $\begin{cases} -3\leq m\leq1 \\ -\sqrt{2}\leq m\leq \sqrt{2}\end{cases}$, solving this gives $-\sqrt{2}\leq m\leq1$.
If $p$ is false and $q$ is true, then $\begin{cases} m1 \\ m\sqrt{2}\end{cases}$, solving this gives $m\sqrt{2}$.
Therefore, the range of the real number $m$ is $m\sqrt{2}$, which can be summarized as $\boxed{m\sqrt{2}}$.