Five. (Total 20 points) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through the fixed point $A(1,0)$, and the 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 the focus, passing through points $B$ and $C$, and opening to the left, with the vertex coordinates $M(m, 0)$. When the eccentricity $e$ of the ellipse satisfies $1>e^{2}>\frac{2}{3}$, find the range of the real number $m$.

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The given ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and it passes through the point $A(1,0)$. The foci of the ellipse are on the $x$-axis. The ellipse intersects the curve $|y|=x$ at points $B$ and $C$. A parabola with $A$ as the focus, passing through points $B$ and $C$, and opening to the left, has the vertex coordinates $M(m, 0)$. When the eccentricity $e$ of the ellipse satisfies $1>e^{2}>\frac{2}{3}$, find the range of the real number $m$.
$$
\begin{array}{l}
a=1, c=\sqrt{1-b^{2}}, e=\sqrt{1-b^{2}} . \\
\because 1>e^{2}>\frac{2}{3}, \therefore \frac{\sqrt{6}}{3}<b<1, \\
\because \frac{p}{2}=m-1, \\
\therefore y^{2}=4(1-m)(x-m) .
\end{array}
$$

Substituting $y=x, x \in\left(0, \frac{1}{2}\right)$ into equation (1) yields
$$
\begin{array}{l}
x^{2}+4(m-1) x-4 m(m-1)=0 \\
\left(m>1,0<x<\frac{1}{2}\right) .
\end{array}
$$

Therefore, $1<m<\frac{3+\sqrt{2}}{4}$.
$$
\begin{array}{l}
\text{In summary, } 1<m<\frac{3+\sqrt{2}}{4}.
\end{array}
$$