Answer:
$$
\frac{8 \sqrt{2}}{9}
$$

Since $A E$ bisects $\angle R A M$, we have $R E=E M$, and $E, A$ lie on different sides of $R M$. Since $A M$ is a diameter, $\angle A R M=90^{\circ}$. If the midpoint of $R M$ is $N$, then from $[R A M]=[R E M]$ and $\angle A R M=90^{\circ}$, we find $A R=N E$. Note that $O$, the center of $\omega, N$, and $E$ are collinear, and by similarity of triangles $N O M$ and RAM, ON $=\frac{1}{2} A R=\frac{1}{2} N E$. Therefore, $O N=\frac{1}{3}$ and $N E=\frac{2}{3}$. By the Pythagorean theorem on triangle $R A M, R M=\frac{4 \sqrt{2}}{3}$, Therefore, the area of $M A R E$ is $2 \cdot \frac{1}{2} \cdot \frac{4 \sqrt{2}}{3} \cdot \frac{2}{3}=\frac{8 \sqrt{2}}{9}$.
5. In the quadrilateral $M A R E$ inscribed in a unit circle $\omega, A M$ is a diameter of $\omega$, and $E$ lies on the angle bisector of $\angle R A M$. Given that triangles $R A M$ and $R E M$ have the same area, find the area of quadrilateral MARE.