3. Given that $P, Q, R, S$ are four points inside the tetrahedron $A-BCD$, and $Q, R, S, P$ are the midpoints of segments $PA, QB, RC, SD$ respectively. Then $V_{P-ABC}: V_{P-BCD}: V_{P-CDA}: V_{P-DAB}=$ $\qquad$
$3.8: 1: 2: 4$.
Let $H_{P, B C D}$ be the distance from point $P$ to plane $B C D$, and so on.
As shown in Figure 3.

Given $V_{P-B C D}=1$.
$$
\begin{array}{l}
\text { By } \frac{H_{S, B C D}}{H_{P, B C D}}=\frac{S D}{P D} \\
=2 \\
\Rightarrow V_{S-B C D}=2 ; \\
\text { By } \frac{H_{R, B C D}}{H_{S, B C D}}=\frac{R C}{S C} \\
\quad=2 \\
\Rightarrow V_{R-B C D}=4 ; \\
\text { By } \frac{H_{Q, B C D}}{H_{R, B C D}}=\frac{Q B}{R B}=2 \Rightarrow V_{Q-B C D}=8 .
\end{array}
$$

Let the extension of $A P$ intersect plane $B C D$ at point $P^{\prime}$. Then
$$
\frac{Q P^{\prime}}{P P^{\prime}}=\frac{V_{Q-B C D}}{V_{P-B C D}}=8 \text {. }
$$

Therefore, $\frac{Q P}{P P^{\prime}}=7$.
Since $A Q=Q P$, we have $\frac{A P^{\prime}}{P P^{\prime}}=15$.
Thus, $V_{A-B C D}=15$.
Similarly, $V_{S-A B C}=1, V_{Q-A C D}=1, V_{R-A B D}=1$.
Therefore, $V_{P-A B C}=8 V_{S-A B C}=8$,
$$
\begin{array}{l}
V_{P-C D A}=2 V_{Q-C D A}=2, \\
V_{P-A B D}=4 V_{R-A B D}=4 .
\end{array}
$$

Hence $V_{P-A B C}: V_{P-B C D}: V_{P-C D A}: V_{P-D A B}$
$$
=8: 1: 2: 4 \text {. }
$$