Five. (Full marks 20 points) As shown in the figure, the base $ABC$ of the tetrahedron $P-ABC$ and the base $\odot O$ of the cone $SO$ are both on the plane $\alpha$, and $\odot O$ passes through point $A$. Furthermore, the diameter $AD \perp BC$ of $\odot O$, with the foot of the perpendicular being $E$. Given that all the edges of the tetrahedron $P-ABC$ are 1, and the diameter and slant height of the cone's base are also 1. Find the distances from the vertex $S$ of the cone to the three lateral faces of the tetrahedron $P-ABC$.
Five, first find the distance from $S$ to the side $P B C$.
Let $E A$ intersect $B C$ at $E$, and connect $P E$, then $P E \perp B C$. Draw $P H \perp A E$, with the foot of the perpendicular at $H$, then $P H \perp \alpha$. It is known that $S O, A E, P E$ are in the same plane $\beta$. Let the intersection point of $P E$ and $S O$ be $M$ (as shown in the figure).

Draw $S N \perp M E$, with the foot of the perpendicular at $N$, then the length of $S \Lambda$ is the distance from $S$ to the side $P B C$.
It is easy to know that $A E = P E = \frac{\sqrt{3}}{2}, E H =$
$$
\begin{array}{l}
\frac{\sqrt{3}}{6}, P H = \frac{\sqrt{6}}{3}, O A = \frac{1}{2}, S O = \\
\frac{\sqrt{3}}{2}, O E = \frac{1}{2}(1 + \sqrt{3}).
\end{array}
$$

From Rt $\triangle P H E \sim R \mathrm{R} \triangle M O E$, we get
$$
\begin{array}{l}
M O = \frac{P H \cdot O E}{E H} = \sqrt{2} + \sqrt{6}. \\
\therefore M S = \sqrt{2} + \sqrt{6} - \frac{\sqrt{3}}{2},
\end{array}
$$
$$
M E = \sqrt{O} \cdot \overline{M^{2} + O E^{2}} = \frac{3}{2}(1 + \sqrt{3})
$$

From Rt $\triangle M N S \sim R t \triangle M C O E$, we get
$$
S N = \frac{M O E}{M E} = \frac{1}{6}(2 \sqrt{2} + 2 \sqrt{6} - \sqrt{3}).
$$

Next, find the distance from $S$ to the side $P A B$.
Connect $C H$ to intersect $A B$ at $F$, and extend it to $G$, such that $O G \parallel A B$ (as shown in the left figure above). Draw $G S^{\prime} \perp$ plane $\alpha$, and take $S^{\prime} G = S O = \frac{\sqrt{3}}{2}$, with $S^{\prime}$ and $S$ on the same side of $\alpha$. Connect $P F$, then $P F \perp A B, A B \perp$ plane $P H F, S^{\prime} G$ is in plane $P H F$. Connect $S^{\prime} S$, then $S^{\prime} S \parallel A B$. Draw $S^{\prime} K \perp P F$, with the foot of the perpendicular at $K$ (as shown in the right figure above), then the length of $S^{\prime} K$ is the distance from $S$ to the side $P A B$.
It is easy to know that $F G = \frac{1}{4}$. Let $\angle F P H = \theta, \angle F S^{\prime} G = \varphi$, then
$$
\operatorname{tg} \theta = \frac{F H}{P H} = \frac{\sqrt{2}}{4}, \operatorname{tg} \varphi = \frac{F G}{S G} = \frac{F G}{S O} = \frac{\sqrt{3}}{6}.
$$

Thus, $\operatorname{tg}(\theta + \varphi) = \frac{6 \sqrt{2} + 4 \sqrt{3}}{24 - \sqrt{6}}$.
$$
\therefore \sin (\theta + \varphi) = \frac{2}{\sqrt{39}} + \frac{4}{3 \sqrt{26}}.
$$

Also, $S^{\prime} F = \sqrt{S^{\prime} G^{2} + F G^{2}} = \frac{\sqrt{13}}{4}$.
In Rt $\triangle S^{\prime} K F$,
$$
S^{\prime} K = S^{\prime} F \sin (\theta + \varphi) = \frac{1}{6}(\sqrt{3} + \sqrt{2}).
$$

By symmetry, the distance from $S$ to the side $P A C$ is also
$$
\frac{1}{6}(\sqrt{3} + \sqrt{2}).
$$