5. As shown in Figure 1, construct sections parallel to the four faces of the regular tetrahedron $P A B C$, such that the four faces of the tetrahedron $P A B C$ are all cut into regular hexagons. After removing the four small tetrahedrons, the resulting polyhedron is denoted as $G$. Then the ratio of the surface area of the tetrahedron $P A B C$ to the polyhedron $G$ is $\qquad$, and the ratio of their volumes is $\qquad$
5. 9: $7,27: 23$.

Let $S, V$ be the area of each face and the volume of the regular tetrahedron, as shown in Figure 4.
According to the given conditions,
$\triangle P A^{\prime} B^{\prime}, \triangle P B^{\prime} C^{\prime},$
$\triangle P C^{\prime} A^{\prime}$ are similar to the triangles on each face of the regular tetrahedron, with a similarity ratio of $1: 3$.
$$
\begin{array}{l}
\text { Then } S_{\triangle P A^{\prime} B^{\prime}}=S_{\triangle P B^{\prime} C^{\prime}}=S_{\triangle P C^{\prime} A^{\prime}} \\
=S_{\triangle A^{\prime} B^{\prime} C^{\prime}}=\frac{1}{9} S .
\end{array}
$$

Therefore, $S_{\text {tetrahedron }}: S_{\text {polyhedron }}$
$$
=4 S:\left(4 S-4 \times 3 \times \frac{1}{9} S+4 \times \frac{1}{9} S\right)=9: 7 \text {. }
$$

And $V_{P-A^{\prime} B^{\prime} C^{\prime}}=\frac{1}{27} V$, thus,
$$
V_{\text {tetrahedron }}: V_{\text {polyhedron }}=V:\left(V-4 \times \frac{1}{27} V\right)=27: 23 \text {. }
$$