Does every null set have a superset which is an $F_{\sigma}$ null set?

The answer, perhaps surprisingly, is no. You write

null sets I had ever seen have been constructed from finite or countable sets or Cantor set

But that doesn't exhaust the full variety of the null sets at all. There are quite a lot of very strange null sets out there, and one of the most important examples (and a counterexample to many seemingly-plausible claims) is the existence of a comeager null set - that is, a null set whose complement is the union of countably many nowhere-dense sets. At first glance such a thing may seem impossible, but they do exist - e.g. take the set of all non-absolutely-normal numbers.

(See e.g. the discussion at this MO question. Basically, category and measure are completely orthogonal, although they're both "notions of size," and the interplay between the two (and other notions of size) gives rise to a lot of interesting analysis, topology, and descriptive set theory.)

By the Baire category theorem, comeager null sets can't be covered "efficiently" by $F_\sigma$ sets. Specifically, BCT implies that (in $\mathbb{R}$) meager $G_\delta$ sets are nowhere dense, so comeager $F_\sigma$ sets contain intervals and hence aren't null. So any comeager null $A$ gives a counterexample to your guess.

A closed set of Lebesgue measure zero has empty interior. So a countable union of such sets, an $F_\sigma$ of measure zero, is meager (also called "first Baire category), and so are all its subsets. But there are Lebesgue null sets that are not meager, for example the set of those numbers in $[0,1]$ whose binary expansion does not have asymptotically half zeros and half ones (i.e., those numbers whose binary expansions violate the strong law of large numbers).