44. For the inequality about $x$, $\left|x-\frac{1}{2}(a+1)^{2}\right| \leqslant \frac{1}{2}(a-1)^{2}$, and $x^{2}-3(a+1) x+2(3 a+1) \leqslant 0$ where $(a \in \mathrm{R})$, the solution sets are $A$ and $B$ respectively, and $A \cap B=A$. Find the range of values for $a$.
44. From the inequality $\left|x-\frac{1}{2}(a+1)^{2}\right| \leqslant \frac{1}{2}(a-1)^{2}$,

we get $A=\left\{x \mid 2 a \leqslant x \leqslant a^{2}+1\right\}$.
From $x^{2}-3(a+1) x+2(3 a+1) \leqslant 0(a \in R)$,
we get $(x-2)[x-(3 a+1)] \leqslant 0$.
When $a \geqslant \frac{1}{3}$, $B=\{x \mid 2 \leqslant x \leqslant 3 a+1\}$;
When $a<\frac{1}{3}$, $B=\{x \mid 3 a+1 \leqslant x \leqslant 2\}$.
Also, $A \cap B=A \Rightarrow A \subseteq B$. When $a \geqslant \frac{1}{3}$, we have $\left\{\begin{array}{l}2 \leqslant 2 a, \\ a^{2}+1 \leqslant 3 a+1\end{array} \Rightarrow 1 \leqslant a \leqslant 3\right.$,

When $a<\frac{1}{3}$, we have $\left\{\begin{array}{l}3 a+1 \leqslant 2 a, \\ a^{2}+1 \leqslant 2\end{array} \Rightarrow a=-1\right.$,
$\therefore$ the range of $a$ is $\{a \mid 1 \leqslant a \leqslant 3$ or $a=-1\}$.