8,9

Point $K$ is located on edge $AD$ of tetrahedron $ABCD$, point $N$ is on the extension of edge $AB$ beyond point $B$, and point $M$ is on the extension of edge $AC$ beyond point $C$, such that $AK:KD=3:1$, $BN=AB$, and $CM:AC=1:3$. Construct the section of the tetrahedron by the plane passing through points $K, M, N$. In what ratio does this plane divide the volume of the tetrahedron?
Let lines $C D$ and $K M$ intersect at point $P$, and lines $B D$ and $K N$ intersect at point $Q$. Consider the plane of face $A C D$. Through point $D$, draw a line parallel to $A C$. Let $T$ be the point of intersection of this line with the extension of $K M$. From the similarity of triangles $D K T$ and $A K M$, we find that

$$
D T = A M \cdot \frac{D K}{A K} = \frac{1}{3} A M = \frac{1}{3} \cdot \frac{4}{3} A C = \frac{4}{3} A C.
$$

From the similarity of triangles $D P T$ and $C P M$, it follows that

$$
\frac{D P}{P C} = \frac{D T}{C M} = \frac{1}{3} A C = \frac{4}{3}
$$

thus, $\frac{D P}{\overline{D C}} = \frac{4}{7}$. Consider the plane of face $A B D$. Through point $D$, draw a line parallel to $A B$. Let $F$ be the point of intersection of this line with the extension of $K N$. From the similarity of triangles $D K F$ and $A K N$, we find that

$$
D F = A N \cdot \frac{D K}{A K} = \frac{1}{3} A N = \frac{1}{3} \cdot 2 A B = \frac{2}{3} A B.
$$

From the similarity of triangles $D Q F$ and $B Q N$, it follows that

$$
\frac{D Q}{Q B} = \frac{D F}{B N} = \frac{2}{3} \cdot \frac{A B}{A B} = \frac{2}{3}
$$

thus, $\frac{D Q}{D B} = \frac{2}{5}$. Therefore,

$$
\frac{V_{K P Q D}}{V_{A B C D}} = \frac{D K}{D A} \cdot \frac{D P}{D C} \cdot \frac{D Q}{D B} = \frac{1}{4} \cdot \frac{4}{7} \cdot \frac{2}{5} = \frac{2}{35}
$$

Let $V_1$ and $V_2$ be the volumes of the polyhedra into which the plane $M N K$ divides the pyramid $A B C D$. Then $\frac{V_1}{V_2} = \frac{2}{33}$.

## Answer

2:33.