Find the end points of the minor and major axis for the graph of the ellipse
\[
\frac{(x-4)^{2}}{9}+\frac{(y-1)^{2}}{36}=1
\]

Highest point on the major axis:
Lowest point on the major axis:

Rightmost point on the minor axis:
Leftmost point on the minor axis:

Highest focal point:
Lowest focal point:

Step 1 ：Given the equation of the ellipse \(\frac{(x-4)^{2}}{9}+\frac{(y-1)^{2}}{36}=1\), we can identify the center of the ellipse as (4,1).

Step 2 ：The length of the semi-major axis is \(\sqrt{36}=6\) and the length of the semi-minor axis is \(\sqrt{9}=3\).

Step 3 ：Since the y-term has the larger denominator, the major axis is vertical. The end points of the major axis are therefore (4, 1-6) and (4, 1+6), which simplify to \(\boxed{(4, -5)}\) and \(\boxed{(4, 7)}\).

Step 4 ：The minor axis is horizontal. The end points of the minor axis are therefore (4-3, 1) and (4+3, 1), which simplify to \(\boxed{(1, 1)}\) and \(\boxed{(7, 1)}\).

Step 5 ：The distance from the center to each focus is given by \(\sqrt{6^2 - 3^2} = 3\sqrt{3}\). Since the major axis is vertical, the foci are located at (4, 1-3\sqrt{3}) and (4, 1+3\sqrt{3}), which are \(\boxed{(4, 1 - 3\sqrt{3})}\) and \(\boxed{(4, 1 + 3\sqrt{3})}\).