2. In the tetrahedron $ABCD$, find a point $P$ such that the volume ratios of the four tetrahedra satisfy $V_{P-BCD}: V_{P-CDA}: V_{P-DAB}: V_{P-ABC} = \alpha: \beta: \gamma: \delta$, where $\alpha, \beta, \gamma, \delta$ are given positive numbers.
2. Take a point $Q$ on the line segment $C D$ such that $C Q: Q D=\delta: r$; take a point $R$ on the line segment $B Q$ such that $B R: R Q=(\gamma+\delta): \beta$; take a point $P$ on the line segment $A R$ such that $A P: P R=(\beta+\gamma+\delta): \alpha$, then point $P$ is the desired point.
In fact, $\frac{V_{P B C D}}{V_{A B C D}}=\frac{R P}{R A}=\frac{\alpha}{\alpha+\beta+\gamma+\delta}$,
$$
\begin{aligned}
\frac{V_{P C D A}}{V_{B C D A}} & =\frac{V_{P C D A}}{V_{R C D A}} \cdot \frac{V_{R C D A}}{V_{B C D A}}=\frac{P A}{R A} \cdot \frac{R Q}{B Q}=\frac{\beta+\gamma+\delta}{\alpha+\beta+\gamma+\delta} \cdot \frac{\beta}{\beta+\gamma+\delta}=\frac{\beta}{\alpha+\beta+\gamma+\delta}, \\
\frac{V_{P D A B}}{V_{C D A B}} & =\frac{V_{P D A B}}{V_{R D A B}} \cdot \frac{V_{R A A B}}{V_{Q D A B}} \cdot \frac{V_{Q D A B}}{V_{C D A B}}=\frac{A P}{A R} \cdot \frac{B R}{B Q} \cdot \frac{D Q}{D C}=\frac{\beta+\gamma+\delta}{\alpha+\beta+\gamma+\delta} \cdot \frac{\gamma+\delta}{\beta+\gamma+\delta} \cdot \frac{\gamma}{\gamma+\delta} \\
& =\frac{\gamma}{\alpha+\beta+\gamma+\delta},
\end{aligned}
$$
$$
\begin{aligned}
\frac{V_{P A B C}}{V_{D A B C}} & =\frac{V_{P A B C}}{V_{R A B C}} \cdot \frac{V_{R A B C}}{V_{Q A B C}} \cdot \frac{V_{Q A B C}}{V_{D A B C}}=\frac{A P}{A R} \cdot \frac{B R}{B Q} \cdot \frac{C Q}{C D}=\frac{\beta+\gamma+\delta}{\alpha+\beta+\gamma+\delta} \cdot \frac{\gamma+\delta}{\beta+\gamma+\delta} \cdot \frac{\delta}{\delta+\gamma} \\
& =\frac{\delta}{\alpha+\beta+\gamma+\delta} .
\end{aligned}
$$

Therefore, $V_{P B C D}: V_{P C D A}: V_{P D 1 B}: V_{P A B C}=\alpha: \beta: \gamma: \delta$, hence point $P$ is the desired point.