3. The three sides of one triangle are $a, a, b$; the three sides of another triangle are $b, b, a (a \neq b)$, and the smallest interior angle of both triangles is equal to $\alpha$. Then $\alpha=$ $\qquad$ $\frac{a}{b}=$ $\qquad$
3. $36 \cdot, \frac{\sqrt{5}-1}{2}$ or $\frac{\sqrt{5}+1}{2}$.

As shown in Figure 6, two triangles are combined to form a quadrilateral $ABCD$, with $AC$ intersecting $BD$ at $K$. It is easy to see that $ABCD$ is an isosceles trapezoid, $\angle A = \angle B = 2\alpha$. At this point, $\angle ADB = 2\alpha$.
From $\triangle ABD$ we know
$$
\alpha + 2\alpha + 2\alpha = 180^{\circ},
$$

then $\alpha = 36^{\circ}$.
It is easy to see that $\triangle KAB \cong \triangle DAC \Rightarrow KC = KD = b - a$.
From $\triangle KAB \sim \triangle KCD \Rightarrow \frac{b}{a} = \frac{a}{b - a}$
$$
\Rightarrow a^2 + ab - b^2 = 0 \text{. }
$$

Solving this, we get $\frac{a}{b} = \frac{\sqrt{5} - 1}{2}$.
By swapping $a$ and $b$, we get $\frac{a}{b} = \frac{2}{\sqrt{5} - 1} = \frac{\sqrt{5} + 1}{2}$.