11. As shown in Figure 1, given
$\odot C:(x+1)^{2}+y^{2}=$
8, a fixed point $A(1,0)$, and $M$ is
a moving point on the circle. Points $P$ and $N$
are on $A M$ and $M C$, respectively, and
satisfy $\overrightarrow{A M}=2 \overrightarrow{A P}, \overrightarrow{N P}$.
$\overline{A M}=0$, the trajectory of point $N$ is curve $E$. If a line passing through the fixed point $F$
$(2,0)$ intersects curve $E$ at two different points $G$ and $H$
(point $G$ is between points $F$ and $H$), and satisfies $\overrightarrow{F G}=\lambda \overrightarrow{F H}$, then the range of $\lambda$ is
11. $[3-2 \sqrt{2}, 1)$.

From the problem, we know that $P$ is the midpoint of $A M$, and $N P \perp A M$, so $N A = N M$.
From $N C + N A = N C + N M = C M = 2 \sqrt{2}$, we get that the locus of point $N$ is an ellipse with foci at $C(-1,0)$ and $A(1,0)$ (as shown in Figure 3).

From $c=1, a=\sqrt{2}$, and $b=1$, we get the equation of curve $E$ as $\frac{x^{2}}{2}+y^{2}=1$.
Thus, $\lambda=\frac{|\overrightarrow{F G}|}{|\overrightarrow{F H}|}$, we have $\frac{2-\sqrt{2}}{2+\sqrt{2}} \leqslant \lambda<1$, which means $3-2 \sqrt{2} \leqslant \lambda<1$.