3. Let $d$ denote the distance between the lines $A B$ and $C D$. Being parallel to $A B$ and $C D$, the plane $\pi$ intersects the faces of the tetrahedron in a parallelogram $E F G H$. Let $X \in A B$ be a points such that $H X \| D B$.

Clearly $V_{A E H B F G}=V_{A X E H}+$ $V_{X E H B F G}$. Let $M N$ be the common perpendicular to lines $A B$ and $C D(M \in A B, N \in C D)$ and let $M N, B N$ meet the plane $\pi$ at $Q$ and $R$ respectively. Then it holds that $B R / R N=M Q / Q N=k$ and consequently $A X / X B=A E / E C=$ $A H / H D=B F / F C=B G / G D=$ k. Now we have $V_{A X E H} / V_{A B C D}=$ $k^{3} /(k+1)^{3}$.
Furthermore, if $h=3 V_{A B C D} / S_{A B C}$ is the height of $A B C D$ from $D$, then $V_{X E H B F G}=\frac{1}{2} S_{X B F E} \frac{k}{k+1} h$ and
$$
S_{X B P E}=S_{A B C}-S_{A X E}-S_{B F C}=\frac{(k+1)^{2}-1-k^{2}}{(k+1)^{2}}=\frac{2 k}{(1+k)^{2}} .
$$

These relations give us $V_{X E H B P G} / V_{A B C D}=3 k^{2} /(1+k)^{3}$. Finally,
$$
\frac{V_{A E H B F G}}{V_{A B C D}}=\frac{k^{3}+3 k^{2}}{(k+1)^{3}} .
$$

Similarly, $V_{C E F D H G} / V_{A B C D}=(3 k+1) /(k+1)^{3}$, and hence the required ratio is $\left(k^{3}+3 k^{2}\right) /(3 k+1)$.
3. (CZS) A tetrahedron $A B C D$ is given. The lengths of the edges $A B$ and $C D$ are $a$ and $b$, respectively, the distance between the lines $A B$ and $C D$ is $d$, and the angle between them is equal to $w$. The tetrahedron is divided into two parts by the plane $\pi$ parallel to the lines $A B$ and $C D$. Calculate the ratio of the volumes of the parts if the ratio between the distances of the plane $\pi$ from $A B$ and $C D$ is equal to $k$.