Solve: From $a^{2} x^{2}+a x-2=0$ we get $(a x+2)(a x-1)=0$.
Obviously $a \neq 0, \therefore x=-\frac{2}{a}$ or $x=\frac{1}{a}$.
Since $x \in[-1,1]$, thus $\left|\frac{2}{a}\right| \leqslant 1$ or $\left|\frac{1}{a}\right| \leqslant 1$, so $|a| \geqslant 1$.
There is only one real number satisfying $x^{2}+2 a x+2 a \leqslant 0$, which means the parabola $y=x^{2}+2 a x+2 a$ intersects the $x$-axis at only one point, so $\Delta=4 a^{2}-8 a=0$, thus $a=0$ or 2.
The proposition “$p$ or $q$” is true when $|a| \geqslant 1$ or $a=0$.
Since the proposition “$p$ or $q$” is false, the range of values for $a$ is $\{a \mid-1<a<0$ or $0<a<1\}$.
Comment: This problem can also be solved by first using a truth table to determine that both $p$ and $q$ are false propositions, and then solving the problem.