17. Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with eccentricity $e=\frac{\sqrt{2}}{2}$, the line $y=2 x-1$ intersects $C$ at points $A, B$, and $|A B|=\frac{8}{9} \sqrt{5}$.
(1) Find the equation of the ellipse $C$;
(2) A line $l$ (with non-zero slope) passing through the point $M(2,0)$ intersects the ellipse $C$ at two different points $E, F$ (with $E$ between $F, M$). Let $\lambda=\frac{S_{\triangle O M E}}{S_{\triangle O M F}}$, find the range of $\lambda$.
(1) From $e=\frac{\sqrt{2}}{2}$, we get $a=\sqrt{2} c=\sqrt{2} b$, so the equation of the ellipse is $x^{2}+2 y^{2}-2 b^{2}=0$.
From $\left\{\begin{array}{l}x^{2}+2 y^{2}-2 b^{2}=0 \\ y=2 x-1\end{array}\right.$, we get $9 x^{2}-8 x+\left(2-2 b^{2}\right)=0$, so $\Delta=64-36\left(2-2 b^{2}\right)$. From $|A B|=\frac{8}{9} \sqrt{5}$, we get $\sqrt{1+2^{2}} \frac{\sqrt{\Delta}}{9}=\frac{8 \sqrt{5}}{9}$, which means $b^{2}=1$, so the equation of the ellipse is $\frac{x^{2}}{2}+y^{2}=1$;
(2) Let $l: x=m y+2$, and $E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right)$, from $\left\{\begin{array}{l}x^{2}+2 y^{2}-2=0 \\ x=m y+2\end{array}\right.$, we get
$$
\left(m^{2}+2\right) y^{2}+4 m y+2=0
$$

So from $\Delta>0$, we solve to get $m^{2}>2$, and $y_{1}+y_{2}=-\frac{4 m}{m^{2}+2}, y_{1} \cdot y_{2}=\frac{2}{m^{2}+2}$
From $\lambda=\frac{S_{\triangle O M E}}{S_{\triangle O M F}}=\frac{\frac{1}{2} \times O M \times y_{1}}{\frac{1}{2} \times O M \times y_{2}}$, we get $y_{1}=\lambda y_{2}$
From (1)(2), we get
$$
\frac{\lambda}{(1+\lambda)^{2}}=\frac{m^{2}+2}{8 m^{2}}=\frac{1}{8}+\frac{1}{4 m^{2}}
$$

So $\frac{1}{8}<\frac{\lambda}{(1+\lambda)^{2}}<\frac{1}{4}$, solving this gives $0<\lambda<3+2 \sqrt{2}$, and $\lambda \neq 1$.