Three, let the height of the regular tetrahedron $P-ABC$ be $PO$, and $M$ be the midpoint of $PO$. A plane passing through $AM$ and parallel to the edge $BC$ divides the tetrahedron into upper and lower parts. Try to find the ratio of the volumes of these two parts.
$\frac{4}{21}$

Three, 【Analysis and Solution】Let the plane through $A M$ parallel to $B C$ intersect the edges $P B$ and $P C$ at $E$ and $F$, respectively, then $E F / / B C$. Also, let $D$ be the midpoint of $B C$, and connect $P D$ to intersect $E F$ at $G$, then
$$
\frac{E F}{B C}=\frac{P G}{T D} \text {. }
$$

Since the distance from $A$ to the plane $P B C$ is the same as the distance from $A$ to the plane $P E F$, we have
$$
\frac{V_{P-A E F}}{V_{P-A B C}}=\frac{S_{\triangle A B C}}{S_{\triangle P B C}}=\left(\frac{P G}{P D}\right)^{2} \text {. }
$$

In $\triangle P D A$, draw a line through $O$ parallel to $P D$ intersecting $A G$ at $N$, then
$$
\frac{P G}{G D}=\frac{O N}{G D}=\frac{A O}{A D}=\frac{2}{3}, \frac{P G}{P D}=\frac{2}{5} \text {, hence } \frac{V_{P-A E F}}{V_{P-A B C}}=\frac{4}{25} \text {. }
$$

Therefore, the ratio of the volumes of the upper and lower parts is $\frac{4}{21}$ (as shown in the figure).