Answer
$$\eqalign{
& {\text{Center}}\left( {3,1} \right) \cr
& {\text{Foci }}\left( {3, - 2} \right){\text{ and }}\left( {3,4} \right) \cr
& {\text{Vertex }}\left( {3, - 4} \right){\text{ and }}\left( {3,6} \right) \cr
& {\text{Eccentricity }}\frac{3}{5} \cr} $$
Work Step by Step
$$\eqalign{
& \frac{{{{\left( {x - 3} \right)}^2}}}{{16}} + \frac{{{{\left( {y - 1} \right)}^2}}}{{25}} = 1 \cr
& {\text{The equation has the standard form}} \cr
& \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1,{\text{ }}a > b \cr
& {\text{Comparing}} \cr
& \frac{{{{\left( {x - 3} \right)}^2}}}{{16}} + \frac{{{{\left( {y - 1} \right)}^2}}}{{25}} = 1 \cr
& \to h = 3,{\text{ }}k = 1,{\text{ }}a = 5,{\text{ }}b = 4,{\text{ }}c = 3 \cr
& {\text{With}} \cr
& \underbrace {{\text{Vertex }}\left( {h,k - a} \right){\text{ and }}\left( {h,k + a} \right)}_ \Downarrow \cr
& {\text{Vertex }}\left( {3,1 - 5} \right){\text{ and }}\left( {3,1 + 5} \right) \cr
& {\text{Vertex }}\left( {3, - 4} \right){\text{ and }}\left( {3,6} \right) \cr
& \underbrace {{\text{Foci }}\left( {h,k - c} \right){\text{ and }}\left( {h,k + c} \right)}_ \Downarrow \cr
& {\text{Foci }}\left( {3,1 - 3} \right){\text{ and }}\left( {3,1 + 3} \right) \cr
& {\text{Foci }}\left( {3, - 2} \right){\text{ and }}\left( {3,4} \right) \cr
& {\text{Eccentricity }}e = \frac{c}{a} = \frac{3}{5} \cr
& {\text{Center}}\left( {h,k} \right) \to {\text{Center}}\left( {3,1} \right) \cr
& {\text{Graph:}} \cr} $$
