Geometry!

The volume of the remaining solid is equal to the volume of the right circular cylinder minus the volume of the right circular cone. We have

$V=V[cylinder]-V[cone]=\pi r^2h - \dfrac{1}{3}\pi r^2 h = \pi (5^2)(12)-\dfrac{1}{3}(\pi)(5^2)(12)=200 \pi$

The total surface area of the remaining solid is equal to the curved surface area of the cone plus the curved surface area of the cylinder plus the area of the base. We have

$A=\pi rL+2\pi r h+ \pi r^2= \pi (5)(13)+ 2(\pi)(5)(12)+ \pi (5^2)=65 \pi + 120 \pi + 25 \pi = 210 \pi$

So the numerical sum is

$V+A= 200 \pi + 210 \pi = 410 \pi = 410 (3.14) = \boxed{1287.4}$