(1) Since point $F(1,0)$ is inside the circle $M: (x+1)^2+y^2=36$, circle $N$ is internally tangent to circle $M$. Therefore, $|NM|+|NF|=6>|FM|$. According to the definition of an ellipse, the trajectory of the center $N$ is an ellipse, with $2a=6$ and $c=1$. Thus, $a^2=9$ and $b^2=8$. Therefore, the equation of the trajectory of the center $N$ of the moving circle is $$\frac{x^2}{9} + \frac{y^2}{8} = 1$$.

(2) Let $P(x_0,y_0)$, $A(x_1,y_1)$, $S(x_S,0)$, and $T(x_T,0)$. Then $B(x_1,-y_1)$. Given that $x_0 \neq \pm x_1$, we have $$k_{AP} = \frac{y_1-y_0}{x_1-x_0}$$, and the equation of line $AP$ is $y-y_1=k_{AP}(x-x_1)$. Setting $y=0$, we get $$x_S= \frac{x_0y_1-x_1y_0}{y_1-y_0}$$, and similarly $$x_T= \frac{x_0(-y_1)-x_1y_0}{(-y_1)-y_0} = \frac{x_0y_1+x_1y_0}{y_1+y_0}$$.
Thus, $$|OS|\cdot|OT|=|x_Sx_T|=| \frac{x_0y_1-x_1y_0}{y_1-y_0} \cdot \frac{x_0y_1+x_1y_0}{y_1+y_0}|=| \frac{x_0^2y_1^2-x_1^2y_0^2}{y_1^2-y_0^2}|$$.
Since $P(x_0,y_0)$ and $A(x_1,y_1)$ are on the ellipse $$\frac{x^2}{9} + \frac{y^2}{8} = 1$$, we have $$y_0^2=8(1- \frac{x_0^2}{9}), y_1^2=8(1- \frac{x_1^2}{9})$$, thus $$y_1^2-y_0^2= \frac{8}{9}(x_0^2-x_1^2), x_0^2y_1^2-x_1^2y_0^2=8x_0^2(1- \frac{x_1^2}{9})-8x_1^2(1- \frac{x_0^2}{9})=8(x_0^2-x_1^2)$$.
Therefore, $$|OS|\cdot|OT|=| \frac{x_0^2y_1^2-x_1^2y_0^2}{y_1^2-y_0^2}|=| \frac{8(x_0^2-x_1^2)}{ \frac{8}{9}(x_0^2-x_1^2)}|=\boxed{9}$$.