$$
B=\left\{x \in \mathbf{R} \mid x^{2}+(2 a-3) x+2 a^{2}-a-3=0\right\} \text {. }
$$

If $A \cup B \neq \varnothing$, find the range of real numbers for $a$.
This problem involves two quadratic equations $x^{2}-(a-2) x-2 a+4=0$ and $x^{2}+(2 a-3) x+2 a^{2}-a-3=0$, at least one of which has real solutions. The direct method is to find the union of the solution sets of the two equations, or the indirect method is to find the complement of the intersection of the sets where the two equations have no solutions.

Solution 1: From the quadratic equation $x^{2}-(a-2) x-2 a+4=0$, we get $\Delta_{1}=(a-2)^{2}-4(-2 a+4)=a^{2}+4 a-12 \geqslant 0 \Rightarrow a \leqslant-6$ or $a \geqslant 2$; From the quadratic equation $x^{2}+(2 a-3) x+2 a^{2}-a-3=0$, we get $\Delta_{2}=(2 a-3)^{2}-4(2 a^{2}-a-3)=-4 a^{2}-8 a+21 \geqslant 0 \Rightarrow -\frac{7}{2} \leqslant a \leqslant \frac{3}{2}$. By $A \cup B \neq \varnothing$, the range of real numbers $a$ is $\{a \mid a \leqslant-6$ or $a \geqslant 2\} \cup \left\{a \left\lvert -\frac{7}{2} \leqslant a \leqslant \frac{3}{2}\right.\right\} = \left\{a \mid a \leqslant-6, -\frac{7}{2} \leqslant a \leqslant \frac{3}{2}\right.$ or $\left.a \geqslant 2\right\}$.

Solution 2: From Solution 1, we have $\left\{\begin{array}{l}\Delta_{1} \geqslant 0 \Rightarrow a \leqslant-6 \text{ or } a \geqslant 2 \\ \Delta_{2} \geqslant 0 \Rightarrow -\frac{7}{2} \leqslant a \leqslant \frac{3}{2}\end{array} \Rightarrow M=\{a \mid -6 < a < -\frac{7}{2} \text{ or } \frac{3}{2} < a < 2\}\right.$. By $A \cup B \neq \varnothing$, the range of real numbers $a$ is $\complement_{\mathrm{R}} M = \left\{a \mid a \leqslant-6, -\frac{7}{2} \leqslant a \leqslant \frac{3}{2}\right.$ or $\left.a \geqslant 2\right\}$.