Solution:
Since A(5, 2) and B(3, 2),
the slope of line AB is $\frac{2-2}{5-3}=0$,
thus, the perpendicular bisector of line AB is vertical to the x-axis, and its equation is: $x=4$,
solving this together with the line $2x-y-3=0$, we get: $x=4$, $y=5$, hence the coordinates of the center of the circle M are (4, 5),
and the radius $r$ of the circle is $r=|AM|=\sqrt{(5-4)^{2}+(2-5)^{2}}=\sqrt{10}$,
therefore, the equation of the circle is $(x-4)^{2}+(y-5)^{2}=10$ $\boxed{(6\ \text{points})}$

(2) Let the equation of the circumcircle of triangle OAB with vertices O(0, 0), A(2, 0), and B(0, 4) be $x^{2}+y^{2}+Dx+Ey+F=0$,
thus, $\begin{cases} F=0 \\ 4+2D+F=0 \\ 16+4E+F=0\end{cases}$,
solving this, we get $D=-2$, $E=-4$, $F=0$,
therefore, the equation of the circumcircle of triangle OAB is $x^{2}+y^{2}-2x-4y=0$ $\boxed{(12\ \text{points})}$