The given equation is $5 y^{2}-9 x^{2}=36$

$\Rightarrow \frac{y^{2}}{\left(\frac{36}{5}\right)}-\frac{x^{2}}{4}=1$ 

$\Rightarrow \frac{y^{2}}{\left(\frac{6}{\sqrt{5}}\right)}-\frac{x^{2}}{2^{2}}=1$    ……….. $(1)$

On comparing equation $( 1 )$ with the standard equation of hyperbola i.e., $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1,$ we obtain $a=\frac{6}{\sqrt{5}}$ and $b=2$

We know that $a^{2}+b^{2}=c^{2}$

$\therefore c^{2}=\frac{36}{5}+4=\frac{56}{5}$

$\Rightarrow c=\sqrt{\frac{56}{5}}=\frac{2 \sqrt{14}}{\sqrt{5}}$

Therefore, the coordinates of the foci are $\left(0,\,\pm \frac{2 \sqrt{14}}{\sqrt{5}}\right)$

The coordinates of the vertices are $\left(0,\,\pm \frac{6}{\sqrt{5}}\right)$

Eccentricity, $e=\frac{c}{a}$ $=\frac{\left(\frac{2 \sqrt{14}}{\sqrt{5}}\right)}{\left(\frac{6}{\sqrt{5}}\right)}$ $=\frac{\sqrt{14}}{3}$

Length of latus rectum $=\frac{2 b^{2}}{a}$ $=\frac{2 \times 4}{\left(\frac{6}{\sqrt{5}}\right)}$ $=\frac{4 \sqrt{5}}{3}$