12. Let the quadratic function $f(x)$ have a positive leading coefficient, and satisfy the inequality
$$
f\left(\frac{3 \pi}{2}-x\right)=f\left(\frac{\pi}{2}+x\right)(x \in \mathbf{R}) \text {. }
$$

Then the solution set of the inequality $f(\arcsin |x|)<f(\arccos |x|)$ is
12. $\left[-1,-\frac{\sqrt{2}}{2}\right) \cup\left(\frac{\sqrt{2}}{2}, 1\right]$.

From the problem, we have
$$
f\left(\pi-\left(x-\frac{\pi}{2}\right)\right)=f\left(\pi+\left(x-\frac{\pi}{2}\right)\right) \text {. }
$$

Thus, the parabola $f(x)$, which opens upwards, has its axis of symmetry at $x=\pi$. Therefore, $f(x)$ is monotonically decreasing for $x \leqslant \pi$.

Given $\arcsin |x|\arccos |x|$, the properties, intersection points, and graphs of the functions $\arcsin |x|$ and $\arccos |x|$ indicate that the solution set of the inequality is
$$
\left[-1,-\frac{\sqrt{2}}{2}\right) \cup\left(\frac{\sqrt{2}}{2}, 1\right] \text {. }
$$