2. Through a vertex of a trapezoid, draw two lines $l_{1}$ and $l_{2}$, where line $l_{1}$ is a diagonal of the trapezoid, and divides the line segment connecting the midpoints of the two bases in the ratio $3: 1$. Line $l_{2}$ bisects the area of the trapezoid. Find the ratio in which line $l_{2}$ divides the line segment connecting the midpoints of the two bases.
2. Let $M$ and $N$ be the midpoints of the bases $BC$ and $AD$, respectively, and $O$ be the intersection of $MN$ and $AC$.
From the problem, we know that the diagonal $AC$ divides $MN$ such that $\frac{MO}{ON}=\frac{1}{3}$.
Similarly, $\frac{BC}{AD}=\frac{MC}{AN}=\frac{1}{3}$.
Let $S$ represent the area of trapezoid $ABCD$.
We will discuss two scenarios below.
(1) If $AL$ bisects the area of the trapezoid, as shown in Figure 1.
Since $S_{\triangle B C}=\frac{1}{3} S_{\triangle A D C}<S_{\triangle D D C}$, point $L$ lies on segment $CD$.
And $E$ is the intersection of line $AL$ and $BC$, then
$S_{\triangle A D L}=\frac{1}{2} S, S_{\triangle A B C}=\frac{1}{4} S, S_{\triangle A C D}=\frac{3}{4} S$.
Note that, $\triangle A C D$ and $\triangle A L D$ have the same height. Therefore,
$$
\begin{array}{l}
\frac{CD}{LD}=\frac{S_{\triangle A C D}}{S_{\triangle A L D}}=\frac{\frac{3}{4}}{\frac{1}{2}}=\frac{3}{2} \\
\Rightarrow \frac{CL}{LD}=\frac{1}{2} \\
\Rightarrow \frac{CE}{AD}=\frac{CL}{DL}=\frac{1}{2} \\
\Rightarrow \frac{MK}{KN}=\frac{ME}{AN}=\frac{MC+CE}{AN} \\
=\frac{MC}{AN}+\frac{CE}{\frac{AD}{2}}=\frac{1}{3}+\frac{1}{1}=\frac{4}{3} .
\end{array}
$$
(2) If $CL$ bisects the area of the trapezoid, as shown in Figure 2.

In this case, point $L$ lies on segment $AD$.
From the problem, we know

Thus, $S_{\triangle K K N}=S_{\triangle C K M}$.
Since $\triangle L K N \sim \triangle C K M$, then $\triangle L K N \cong \triangle C K M$.
Therefore, $\frac{MK}{KN}=1$.
In summary, $\frac{MK}{KN}=\frac{3}{4}$ or 1.