3. Set $A=\left\{x \mid \sqrt{a^{2}+4 x-x^{2}}>a(x+1)\right\}, B=$ $\left\{\left.x|| x-\frac{1}{2} \right\rvert\,<\frac{1}{2}\right\}$. When $A \subseteq B$, the range of values for $a$ is
$3.1 \leqslant a \leqslant \sqrt{2}$.
$$
\because\left|x-\frac{1}{2}\right|<\frac{1}{2}, \therefore 0<x<1 \text {. }
$$

Let $y=a(x+1), y=\sqrt{a^{2}+4 x-x^{2}}$, then
$$
(x-2)^{2}+y^{2}=a^{2}+4(y \geqslant 0) \text {. }
$$

As shown in Figure 3, it is easy to find the intersection points of the two curves are $E(0, a)$,
$$
F\left(\frac{4-2 a^{2}}{a^{2}+1}, \frac{a\left(5-a^{2}\right)}{a^{2}+1}\right)
$$

When $A \subseteq B$, by $B(0,1)$, we have $0<\frac{4-2 a^{2}}{a^{2}+1}$ $\leqslant 1$, then $1 \leqslant a<\sqrt{2}$.

Furthermore, when the line $y=a(x+1)$ is tangent to the semicircle, $a=\sqrt{2}$. At this time, $A=\varnothing$ still satisfies $A \subseteq B$.
In summary, $1 \leqslant a \leqslant \sqrt{2}$.