45. Given the proposition $p$ : the equation $a^{2} x^{2}+a x-2=0$ has a solution in $[-1$, 1]; proposition $q$ : there is only one real number $x$ that satisfies the inequality $x^{2} + 2 a x + 2 a \leqslant 0$, if the proposition " $p$ or $q$ " is false, find the range of values for $a$.
45. 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 $x=\frac{1}{a}$.
$\because x \in[-1,1]$, hence $\left|\frac{2}{a}\right| \leqslant 1$ or $\left|\frac{1}{a}\right| \leqslant 1, \therefore|a|$ $\geqslant 1$. "There is only one real number satisfying $x^{2}+2 a x+2 a \leqslant 0$". That is, the parabola $y=x^{2}+2 a x+2 a$ intersects the $x$-axis at only one point, $\therefore \Delta=4 a^{2}$ $-8 a=0, \therefore a=0$ or 2, so the proposition " $p$ or $q$ is true" when, " $|a| \geqslant 1$ or $a=0$ ".
$\because$ The proposition " $p$ or $q$ " is false,
$\therefore$ the range of values for $a$ is $\{a \mid-1<a<0$ or $0<a<1\}$.