electronic configuration - What is meant by complete outer shell? Why do the noble gases have zero valency?

Does having 8 or 2 electron in the outmost shell mean its outmost shell is full and its valency is zero?

I know that the 3rd and 4th shell can contain 18 and 32 electrons. Then how can Argon's (2,8,8) outmost shell be full, though it does not contain the highest number of electron (18) in it's 3rd shell?

It is the same in the cases of other noble gases too? Why?

Answer

When we refer to "outer shells" we are talking about the highest n-level, but also limiting ourselves to the s- and p-orbitals. The reason for this has to do with how the effective nuclear charge "felt" by the valence electrons changes as you move through successively higher energy configurations.

In summary, $s$- and $p$- electrons are screened less effectively by inner shells, and so for a given n-value, the $s$ and $p$ orbitals of the next n fill before the current n (they have lower energy).

In your example using argon, this means that the 3d shell actually has no electrons - the condensed configuration is:

$$ \ce{Ar: \space [Ne]}3s^23p^6 $$

For the next noble gas, krypton, the condensed electron configuration is:

$$ \ce{Kr: \space [Ar]}4s^23d^{10}4p^6 $$

Note that it has $3d$ electrons at a higher energy than what argon had, but since the value of n roughly corresponds to the distance of the orbital from the nucleus, the $3d$ orbitals are physically closer to the nucleus than the $4s$ and $4p$ orbitals. In other words, for single atoms, the lowest energy electron configuration always fills the s- and p- orbitals first, and so when we are looking for the valence or "outermost" electrons, that is where we need to look - we sort of ignore the $d$- and $f$-orbital electrons for these kinds of problems.