Example 4 The inequality about the real number $x$ is $\left|x-\frac{(a+1)^{2}}{2}\right| \leqslant \frac{(a-1)^{2}}{2}$ and $x^{2}-3(a+1) x+$ $2(3 a+1) \leqslant 0$ (where $a \in \mathbf{R})$. The solution sets of these inequalities are denoted as $A$ and $B$ respectively. Find the range of values for $a$ that makes $A \subseteq B$.
Solve: From $\left|x-\frac{(a+1)^{2}}{2}\right| \leqslant \frac{(a-1)^{2}}{2}$, we get $A=\left\{x \mid 2 a \leqslant x \leqslant a^{2}+1\right\}$; from $x^{2}-3(a+1) x+2(3 a+1) \leqslant 0$, i.e., $(x-2)[x-(3 a+1)] \leqslant 0$.
When $3 a+1 \geqslant 2$ i.e., $a \geqslant \frac{1}{3}$, $B=\{x \mid 2 \leqslant x \leqslant 3 a+1\}$;
When $3 a+1<2$ i.e., $a<\frac{1}{3}$, $B=\{x \mid 3 a+1 \leqslant x \leqslant 2\}$;
When $a \geqslant \frac{1}{3}$, from $A \subseteq B$ we get $\left\{\begin{array}{l}2 \leqslant 2 a \\ a^{2}+1 \leqslant 3 a+1,\end{array}\right.$ so $1 \leqslant a \leqslant 3$;
When $a<\frac{1}{3}$, from $A \subseteq B$ we get $\left\{\begin{array}{l}3 a+1 \leqslant 2 a \\ a^{2}+1 \leqslant 2,\end{array}\right.$ so $a=-1$.

Therefore, the range of $a$ is $\{a \mid 1 \leqslant a \leqslant 3$ or $a=-1\}$.