 Can 3 lights be placed on the outside of any convex N dimensional solid so that all points on its surface are illuminated? 
Already in three dimensions, the answer is no.

Let the solid be a sphere. Consider two lights in arbitrary positions. Together with the center of the sphere, they define a plane. The two points on the sphere farthest from the plane are antipodal, and are not illuminated by the two lights. The third light cannot illuminate both antipodal points.



The proof generalizes readily to $n$ dimensions. It is insufficient to use $n$ lights, because $n-1$ of them along with the center of the hypersphere define an $(n-1)$-dimensional hyperplane, which yields two dark antipodes, and you know the rest.

Also, $n+1$ lights should be sufficient for any convex body: place them at the vertices of a simplex that strictly contains the body. (A rigorous justification escapes me at the moment, but it intuitively seems very clear.) In particular, this answers @Listing's comment in the affirmative.
