Solution:

The answer is (C). Triangle $AHM$ is right-angled, so we have $\overline{HN} = \overline{AM} / 2$ because $HN$ is the median relative to the hypotenuse. Setting $\overline{BH} = x$ we have $\overline{HC} = \overline{BC} - \overline{BH} = 14 - x$ and, by the Pythagorean theorem, $\overline{AH}^2 = \overline{AB}^2 - \overline{BH}^2 = \overline{AC}^2 - \overline{HC}^2$, that is $13^2 - x^2 = 15^2 - (14 - x)^2$, from which we obtain $x = 5$. We then have
$$\overline{AH} = \sqrt{13^2 - 5^2} = 12, \quad \overline{AM} = \sqrt{\overline{AH}^2 + \overline{HM}^2} = \sqrt{12^2 + 2^2} = \sqrt{148} = 2\sqrt{37},$$
so $\overline{HN} = \sqrt{37}$.