II. (25 points) As shown in Figure 10, the distance between two parallel lines $l_{1}$ and $l_{2}$ is 6. There is a fixed circle $\odot O$ with a radius of 1 between $l_{1}$ and $l_{2}$, tangent to line $l_{2}$ at point $A$. $P$ is a moving point on line $l_{1}$. Two tangents $PB$ and $PC$ are drawn from $P$ to $\odot O$, with the points of tangency being $B$ and $C$, respectively, intersecting line $l_{2}$ at points $M$ and $N$. Is $AM \cdot AN$ a constant value? If so, find this constant value; if not, explain the reason.
As shown in Figure 15, draw $PD \perp l_{2}$ at point $D$, and connect $OA, OB, OC, OM, ON,$ and $OP$. Then
$$
\begin{array}{l}
PD=6, \\
OA=OB=OC=1 .
\end{array}
$$

Let $AM=m, AN=n, PC=p, DN=x$, then
$$
DM=m+n-x .
$$

From the problem, $\odot O$ is the incircle of $\triangle PMN$, so
$$
\begin{array}{l}
BM=AM=m, CN=AN=n, PB=PC=p ; \\
OA \perp MN, OB \perp PM, OC \perp PN . \\
\text { Also, } S_{\triangle PMN}=S_{\triangle OMN}+S_{\triangle ONP}+S_{\triangle OPM} \\
=\frac{1}{2} MN \cdot OA+\frac{1}{2} PN \cdot OC+\frac{1}{2} PM \cdot OB \\
=\frac{1}{2}(MN+PN+PM)=m+n+p, \\
S_{\triangle PMN}=\frac{1}{2} MN \cdot PD=3(m+n),
\end{array}
$$

Thus, $m+n+p=3(m+n)$.
Hence, $p=2m+2n$.
In $\mathrm{Rt} \triangle PDN$ and $\mathrm{Rt} \triangle PDM$, by the Pythagorean theorem, we have
$$
\left\{\begin{array}{l}
PD^{2}+DN^{2}=PN^{2}, \\
PD^{2}+MD^{2}=PM^{2},
\end{array}\right.
$$

which gives
$$
\left\{\begin{array}{l}
6^{2}+x^{2}=(p+n)^{2}, \\
6^{2}+(m+n-x)^{2}=(m+p)^{2} .
\end{array}\right.
$$
(2) - (3) yields
$$
\begin{array}{l}
(m+n)(2x-m-n)=(n-m)(m+n+2p) \\
=5(n-m)(m+n),
\end{array}
$$

which simplifies to
$$
x=3n-2m .
$$

Substituting equation (4) into equation (2) gives
$$
36+(3n-2m)^{2}=(2m+3n)^{2} ,
$$

which simplifies to $36=24mn$.
Thus, $mn=1.5$, i.e., $AM \cdot AN=1.5$.
Therefore, $AM \cdot AN$ is a constant, and the constant value is 1.5.