With respect to (a) first:
Why isn't $\bar0$ defined? It seems you're working in $\mathbb{Z}/p\mathbb{Z}$ and not only the group of units; thus $\bar{0}$ is defined just fine. Secondly, it seems you are deliberately forgetting that $\mathbb{Z}/p\mathbb{Z}$ is a ring and not just a group, i.e. that you have both addition and multiplication. Perhaps you thought in both cases that you were working in the group of units instead of the ring of integers mod $p$?
What you should do is work in the ring, realize $a$ has a (non-zero unital) inverse, and conclude that $a$ is in the group of units. This angle is the one espoused by (a), and is what makes it fine.
With respect to (b):
I like this one a lot less because standard proofs of Fermat's little theorem use (often explicitly) the size of the group of units (and Lagrange's theorem). Or they use the size of the group of units and then multiply them all together, do a little cancellation (maybe with Wilson's theorem), and get the result. But in both cases, you use the size of the group of units, so there is danger of self-referentiality.
But there are ways of proving Fermat's little theorem avoiding knowing the size of the group of units; so let's assume we've done that.
A priori, you only know that $a^{p-2} \in \mathbb{Z}/p\mathbb{Z}$, i.e. that it could be $0$. Well, is it $0$? If it were $0$, this means that $p \mid a^{p-2}$. But $p$ is prime, so if $p \nmid a$ then $p \nmid a^{x}$. So $a^{p-2} \neq 0$, and thus is in $1, \ldots, p-1$. (I stopped using bars, but every number here is in $\mathbb{Z}/p\mathbb{Z}$).
The video states that "the set of all cosets forms a group". This is exactly the set $\mathbb{Z}/n\mathbb{Z}$ as you know it. It doesn't say that a coset itself is a subgroup of $\mathbb{Z}$. And indeed, the coset $3 + 5\mathbb{Z}$ is not a subgroup of $\mathbb{Z}$, as you noted. The set $3 + 5\mathbb{Z}$ could be turned into a group though, if we define some silly operation on it that has the identity element $-12$ for example.
Best Answer
To the first question, just quoting your intro, "$a,b\in(\mathbb{Z}/n\mathbb{Z})^{\times}$". So if $\mathbb{Z}/n\mathbb{Z}$ is a set of equivalence classes, then yes, $a$ and $b$ are equivalence classes.
To the second question, $a^{-1}$ is an equivalence class such that $aa^{-1}$ is the equivalence class of $1$. In general, there is no quick formula to find an element of $a^{-1}$ given $a$. Instead, since a representative $A$ of $a$ and $n$ are relatively prime, the Euclidean algorithm provides solutions to the equation $$AB + nt = 1$$ Then mod $n$, $AB\equiv 1$. So the Euclidean algorithm will lead you to a representative of $a^{-1}$.
Now, to back-peddle a little bit, actually there is a rather simple formula for a representative of $a^{-1}$, given a representative $A$ of $a$. Take $B=A^{\varphi(n)-1}$, where $\varphi$ is Euler's totient function. Then $AB=A^{\varphi(n)}\equiv1\mod{n}$. The problem with this "simple" formula is that it's not computationally efficient. For example, if $n=97$, the Euclidean algorithm quickly tells us that $2^{-1}=49$. But this formula would give us $2^{96}$, which is tiresome to compute even if we reduce at every multiplication.