Answer
center $(0,0)$, transverse axis $y$-axis, vertices $(0,-2),(0,2)$, foci $(0,-2\sqrt {5}),(0,2\sqrt {5})$, and asymptotes $y= \pm\frac{1}{2}x$.
See graph.
Work Step by Step
1. Given $4y^2-x^2=16\Longrightarrow \frac{y^2}{4}-\frac{x^2}{16}=1$, we can find the center $(0,0)$, transverse axis $y$-axis, $a=2,b=4$, vertices $(0,-2),(0,2)$, $c=\sqrt {a^2+b^2}=\sqrt {4+16}=2\sqrt {5}$, foci $(0,-2\sqrt {5}),(0,2\sqrt {5})$, and asymptotes $y=\pm\frac{a}{b}x=\pm\frac{1}{2}x$.
2. See graph.
