Answer
$$\eqalign{
& {\text{Center }}\left( {0,0} \right) \cr
& {\text{Vertices}}:\left( {0, \pm 8} \right) \cr
& {\text{Foci: }}\left( {0, \pm 2\sqrt {17} } \right) \cr
& {\text{Asymptotes: }}y = \pm \frac{5}{7}x \cr
& {\text{domain: }}\left( { - \infty , + \infty } \right) \cr
& {\text{range: }}\left( { - \infty ,8} \right] \cup \left[ {8,\infty } \right) \cr} $$
Work Step by Step
$$\eqalign{
& \frac{{{y^2}}}{{64}} - \frac{{{x^2}}}{4} = 1 \cr
& {\text{The equation is written in the form }}\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1 \cr
& \frac{{{y^2}}}{{64}} - \frac{{{x^2}}}{4} = 1,{\text{ then }}a = 8,\,\,b = 2 \cr
& c = \sqrt {{a^2} + {b^2}} = \sqrt {64 + 4} = 2\sqrt {17} \cr
& \cr
& {\text{Therefore,}} \cr
& {\text{Center }}\left( {0,0} \right) \cr
& {\text{Vertices: }}\left( {0, \pm a} \right):\left( {0, \pm 8} \right) \cr
& {\text{Foci: }}\left( {0, \pm c} \right):\left( {0, \pm 2\sqrt {17} } \right) \cr
& {\text{Asymptotes: }}y = \pm \frac{a}{b}x \cr
& {\text{Asymptotes: }}y = \pm 4x \cr
& {\text{The domain of the hyperbola is }}\left( { - \infty , + \infty } \right) \cr
& {\text{The range of the hyperbola is: }}\left( { - \infty ,a} \right] \cup \left[ {a,\infty } \right) \cr
& {\text{range: }}\left( { - \infty ,8} \right] \cup \left[ {8,\infty } \right) \cr
& \cr
& {\text{Graph}} \cr} $$
