2. Fold a square piece of paper to form a triangle and a quadrilateral. It is known that the ratio of their areas is $3: 5$. Then the ratio of their perimeters is
2. $\frac{6}{7}$.

As shown in Figure 5, $ABCD$ is a square, and point $M$ is on side $AB$. Let the side length of the square be $1$, $BM=x$, then $AM=1-x$. Therefore,
$$
\frac{3}{5}=\frac{S_{\triangle ADM}}{S_{\text{quadrilateral } BCDM}}=\frac{\frac{1}{2} AM \cdot AD}{\frac{1}{2}(BM+DC) \cdot BC}=\frac{1-x}{1+x} \text{.}
$$

Solving for $x$ gives $x=\frac{1}{4}$. At this point, $DM=\frac{5}{4}$, thus,
$$
\frac{\text{Perimeter of } \triangle ADM}{\text{Perimeter of quadrilateral } BCDM}=\frac{1+\frac{3}{4}+\frac{5}{4}}{\frac{1}{4}+1+1+\frac{5}{4}}=\frac{6}{7} \text{.}
$$