13. (20 points) Let the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ pass through the point $P\left(\frac{\sqrt{6}}{2}, \frac{1}{2}\right)$, with an eccentricity of $\frac{\sqrt{2}}{2}$, and let the moving point $M(2, t)$ $(t>0)$.
(1) Find the standard equation of the ellipse.
(2) Find the equation of the circle with diameter $OM$ that is intersected by the line $3 x-4 y-5=0$ to form a chord of length 2.
(3) Let $F$ be the right focus of the ellipse. Draw a perpendicular from point $F$ to $OM$ that intersects the circle with diameter $OM$ at point $N$. Prove that the length of segment $ON$ is a constant, and find this constant.
13. (1) From the problem, we know $\frac{\frac{6}{4}}{a^{2}}+\frac{\frac{1}{4}}{b^{2}}=1$. Then $\frac{1}{2}=\frac{c^{2}}{a^{2}}=\frac{a^{2}-b^{2}}{a^{2}}$.
Solving simultaneously, we get $a^{2}=2, b^{2}=1$.
Thus, the equation of the ellipse is $\frac{x^{2}}{2}+y^{2}=1$.
(2) It is easy to see that the midpoint of $O M$ is $\left(1, \frac{t}{2}\right)$.
Then $\left(\frac{\left|3 \times 1-4 \times \frac{t}{2}-5\right|}{\sqrt{3^{2}+(-4)^{2}}}\right)^{2}+1^{2}=1^{2}+\left(\frac{t}{2}\right)^{2}$ $\Rightarrow t=4$ or $t=-\frac{4}{9}$ (discard).
Thus, the equation of the circle is
$$
(x-1)^{2}+(y-2)^{2}=5 \text {. }
$$
(3) Let $N\left(x_{0}, y_{0}\right)$.
Since point $N$ lies on the circle with $O M$ as its diameter, we have $\overrightarrow{O N} \cdot \overrightarrow{M N}=0$.
Then $x_{0}\left(x_{0}-2\right)+y_{0}\left(y_{0}-t\right)=0$
$\Rightarrow x_{0}^{2}+y_{0}^{2}=2 x_{0}+t y_{0}$.
Also, $\overrightarrow{O M} \cdot \overrightarrow{F N}=0$, giving
$2\left(x_{0}-1\right)+t y_{0}=0 \Rightarrow 2 x_{0}+t y_{0}=2$.
Thus, $O N=\sqrt{x_{0}^{2}+y_{0}^{2}}$
$=\sqrt{2 x_{0}+t y_{0}}=\sqrt{2}$ (a constant).