Given: Triangular pyramid $S-A B C$, the base is an equilateral triangle with side length $4 \sqrt{2}$, the edge $S C$ has a length of 2 and is perpendicular to the base, $E$ and $D$ are the midpoints of $B C$ and $A B$ respectively. Find: The distance between $C D$ and $S E$.

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[3D Geometry Solution] Since $CD$ and $SE'$ are two skew lines, it is not obvious where their common perpendicular intersects these two lines. Therefore, the problem needs to be transformed into whether a plane can be found that passes through $SE$ and is parallel to $CD$. $E$ is the midpoint of $BC$. Find the midpoint $F$ on $BD$, and connect $EF$. $EF$ is the midline, so $EF \parallel CD$, hence $CD \parallel$ plane $SEF$. Then,  
the distance from $CD$ to plane $SEF$ is the distance between the two skew lines (by Theorem 1). The distance between a line and a plane can also be converted to the distance from a point $C$ on the line to the plane $SEF$. Let $CG$ be the distance from point $C$ to plane $SEF$.

Connect $CF$, using the fact that the volumes of tetrahedra $S-CFE$ and $C-EFS$ are the same, we can find the length of $CG$.

Given $BC = 4\sqrt{2}$, $E$ and $F$ are the midpoints of $BC$ and $BD$, respectively, we can get $CD = 2\sqrt{6}$, $EF = \sqrt{6}$, $DF = \sqrt{2}$. Also, $SC = 2$, so
\[
V_{C-EFS} = \frac{1}{3} \left( \frac{1}{2} \sqrt{6} \cdot \sqrt{2} \right) \cdot 2 = \frac{2\sqrt{3}}{3}.
\]
On the other hand,
\[
\text{Let } a = SE - EFS = \frac{1}{3} S_{\triangle BFS} \cdot CG.
\]
\[
b = \sqrt{4 + 8} = \sqrt{2} \cdot \sqrt{6},
\]
\[
c = EF = \sqrt{4 + 26} = \sqrt{2} \cdot \sqrt{15},
\]
\[
s = \frac{1}{2}(a + b + c) = \frac{\sqrt{2}}{2}(\sqrt{3}).
\]

Using Heron's formula to find the area of $\triangle EFS$:
\[
S_{\triangle EFS}^2 = s(s - a)(s - b)(s - c)
\]
\[
= \frac{1}{4} \left[ (\sqrt{3} + \sqrt{15})^2 - 3 \right] \cdot \left[ 3 - (\sqrt{6} - \sqrt{15})^2 \right]
\]
\[
= \frac{1}{4} (6 + 15 + 2\sqrt{90} - 3) \cdot (3 - 6 - 15 + 2\sqrt{90})
\]
\[
= \frac{1}{4} (6\sqrt{10} + 18)(6\sqrt{10} - 18)
\]
\[
= y.
\]
\[
V_{C-EFS} = \frac{1}{3} \cdot 3 \cdot CG = CG.
\]
\[
\because V_{C-EFS} = V_{S-GPE},
\]
\[
\frac{2\sqrt{3}}{3}.
\]

Therefore, $\triangle EFS = 3$.
\[
\therefore CG = \frac{2\sqrt{3}}{3}, \text{ i.e., the distance between } CD \text{ and } SE \text{ is } \frac{2\sqrt{3}}{3}.
\]

[Analytical Method]
Establish a spatial rectangular coordinate system, with $C$ as the origin, the line containing $CD$ as the $x$-axis, and the line containing $SC$ as the $z$-axis (as shown in the figure). Use the midpoints $E$ and $F$ of $BC$ and $BD$.

To find the distance from $C$ to the plane $SEF$, we can use the analytical method and the point-to-plane distance formula. From the given conditions, we determine: $C(0,0,0)$, $S(0,0,2)$, $F(2\sqrt{6}, \sqrt{2}, 0)$, $E'(\sqrt{6}, \sqrt{2}, 0)$. The plane equation determined by points $S$, $E$, and $F$ is:
\[
\left| \begin{array}{ccc}
x - 0 & y - 0 & z - 2 \\
\sqrt{6} & \sqrt{2} & -2 \\
2\sqrt{6} & \sqrt{2} & 2
\end{array} \right| = 0,
\]
which simplifies to
\[
2y + \sqrt{2}z - 2\sqrt{2} = 0.
\]
Let the distance from $C$ to the plane $SEF$ be $d$, then
```
[Algebraic Method] Let the common perpendicular segment of $SE$ and $CD$ be $FG$ (as shown in the figure). Draw a perpendicular from $F$ to $GH$, then
$GH \parallel SO$. Connect $IH$.
    Since $OSC \perp$ plane $ABC$,
    and $HF \subset$ plane $ABC$,
    we know $\angle BCD = 30^\circ$. Let $FH = x$, then $CH = 2x$. Therefore,
    \[
    FG^2 = FH^2 + GH^2 = x^2 + (2 - \sqrt{2}x)^2 = 3x^2.
    \]
    \[
    SF^2 = SC^2 + CF^2 = 4 + (\sqrt{3}x)^2 = 3x^2.
    \]
    \[
    SE = \sqrt{4 + (2\sqrt{2})^2} = \sqrt{12}.
    \]
    \[
    SG = \sqrt{12} \cdot \frac{2x}{2\sqrt{2}} = \sqrt{6}x.
    \]
    From (1) and (2),
    \[
    3x^2 = 3x^2.
    \]
    Solving for $x$,
    \[
    x = \frac{2\sqrt{2}}{3}.
    \]
```