Five, it is known that the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through the fixed point $A(1, 0)$, and its foci are on the $x$-axis. The ellipse intersects the curve $|y|=x$ at points $B$ and $C$. There is a parabola with $A$ as its focus, passing through points $B$ and $C$, and opening to the left, with the vertex coordinates of the parabola being $M(m, 0)$. When the eccentricity $e$ of the ellipse satisfies $\frac{2}{3}<e^{2}<1$, find the range of the real number $m$.
Five, the ellipse passes through the fixed point $A(1,0)$, then
$a=1, c=\sqrt{1-b^{2}}, e=\sqrt{1-b^{2}}$.
Since $\frac{2}{3}1,00, m>1) \text {. }
$$

Since $\frac{p}{2}=m-1$, then,
$$
\left\{\begin{array}{l}
f(0)=-4 m(m-1)0 .
\end{array}\right.
$$

Therefore, $\left\{\begin{array}{l}m>1 \text { or } m1) \text {. }
$$
(Provided by Liu Jinguo)