Since the lines $BC$ and $DE$ are parallel and since $M$ and $N$ are mid-points of the line segments $DB$ and $EC$, respectively, it is easy to see that the line $MN$ is also parallel to the line $BC$ (and $DE$). Therefore, the triangles $ADE$, $AMN$ and $ABC$ are similar. If we let $a = DE$, $b = BC$ and let $h$ and $\ell$ be the height of the triangles $ADE$ and $ABC$, respectively, then we have $MN = \frac{1}{2}(a+b)$ and the height of the trapezoid $DMNE$ (and the trapezoid $MBCN$) equals $\frac{1}{2}(\ell-h)$. Since the areas of the trapezoids $DMNE$ and $MBCN$ are $1$ and $2$, respectively, we have

$$\frac{1}{2} \left( a + \frac{a+b}{2} \right) \cdot \left( \frac{\ell - h}{2} \right) = 1 \quad \text{and} \quad \frac{1}{2} \left( b + \frac{a+b}{2} \right) \cdot \left( \frac{\ell - h}{2} \right) = 2,$$

from which we obtain $b = 5a$. From the similarity of the triangles $ADE$ and $ABC$, we also get $\ell = 5h$. The area $S_1$ of the triangle $ADE$ is $\frac{1}{2}ah$ and the area $S_2$ of the triangle $ABC$ is $\frac{1}{2}bh = \frac{25}{2}ah$, and since we also have
$$S_2 - S_1 = \text{the area of the trapezoid } DBCE = 1 + 2 = 3,$$
we get
$$\frac{25}{2}ah - \frac{1}{2}ah = 3,$$
from which it follows that $ah = \frac{1}{4}$ and we obtain the desired answer $S_1 = \frac{1}{2}ah = \frac{1}{8}$.