15. Let the inequality about $x$ be $\left|x-\frac{(a+1)^{2}}{2}\right| \leqslant \frac{(a-1)^{2}}{2}$ 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 $A$ and $B$, respectively. If $A \subseteq B$, find the range of values for $a$.
15. From $\left|x-\frac{(a+1)^{2}}{2}\right| \leqslant \frac{(a-1)^{2}}{2}$, we get $2 a \leqslant x \leqslant a^{2}+1$, so $A=\left[2 a, a^{2}+1\right]$. From $x^{2}-$ $3(a+1) x+2(3 a+1) \leqslant 0$, we get $(x-2)(x-3 a-1) \leqslant 0$.
(1) If $a \geqslant \frac{1}{3}$, then $B=[2,3 a+1]$. By $A \subseteq B$ we get $\left[2 a, a^{2}+1\right] \subseteq[2,3 a+1]$, then
$$
\left\{\begin{array}{l}
a \geqslant \frac{1}{3} \\
2 a \geqslant 2 \\
a^{2}+1 \leqslant 3 a+1
\end{array}\right.
$$
(2) If $a<\frac{1}{3}$, then $B=[3 a+1,2]$. By $A \subseteq B$ we get $\left\{\begin{array}{l}a<\frac{1}{3} \\ 2 a \geqslant 3 a+1 \\ a^{2}+1 \leqslant 2,\end{array}\right.$ which gives $a=-1$.

In summary, $a \in[1,3] \cup\{-1\}$.