Given the ellipse $\frac{(x-4)^{2}}{4}+\frac{(y-3)^{2}}{36}=1$ Find the center point: List the vertices:

Step 1 ：Given the ellipse equation \(\frac{(x-4)^{2}}{4}+\frac{(y-3)^{2}}{36}=1\)

Step 2 ：The center of the ellipse is given by the coordinates (h, k) where h and k are the values in the equation. In this case, the center is (4, 3)

Step 3 ：The vertices of the ellipse are given by the points (h±a, k) and (h, k±b) where a and b are the square roots of the denominators in the equation. In this case, the vertices are (2,3), (6,3), (4,-3), (4,9)

Step 4 ：Final Answer: The center of the ellipse is \(\boxed{(4,3)}\) and the vertices are \(\boxed{(2,3), (6,3), (4,-3), (4,9)}\)