Galois connection for annhilators

I think you must have typoed the first one. I think you rather mean

$ann(ann(ann(S))) = ann(S)$

because, for example, if you take the zero ideal in $mathbb Z$, your first equation is wrong as written.

I'm going to use some ad hoc notation to write the general case.

Let $M$ be a right $R$ module and write

$ann(X)={rin Rmid Xr={0}}$ for a nonempty subset $Xsubseteq M$

$Ann(Y)={min Mmid mY={0}}$ for a nonempty subset $Ysubseteq R$

Now note:

1. Just by the definitions of annihilators, $Ysubseteq ann(Ann(Y))$ and $Xsubseteq Ann(ann(X))$.

2. Furthermore, it is easy to check that both maps $ann()$ and $Ann$ are containment reversing.

Now, using the first equation, substituting $ann(X)$ for $Y$, you'd get $ann(X)subseteq ann(Ann(ann(X)))$. On the other hand, applying $ann$ to both sides of $Xsubseteq Ann(ann(X))$, you get $ann(X)supseteq ann(Ann(ann(X))$. Therefore $ann(X)=ann(Ann(ann(X)))$.

Using a similar argument, you have $Ann(Y)=Ann(ann(Ann(Y)))$.

From this, you can derive that the left and right annihilator maps in $R$ satisfy $ell r ell = ell$ and $r ell r=r$, and in the case of a commutative ring you'd have the identity you proposed above: $ann^3=ann$.

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As for the second part, from $Ysubseteq ann(Ann(Y))$ you have that the left ideal generated by $Y$ is contained in $ann(Ann(Y))$ since $ann(Ann(Y))$ is a left ideal of $R$.

And likewise, you have that the submodule generated by $X$ is contained in $Ann(ann(X))$ since $Ann(ann(X))$ is a submodule of $M$.

But in general it does not have to be the case that you have equality.

You can work out why $ann(Ann(L))=L$ if and only if $L$ is of the form $ann(N)$ for some submodule $N$ of $M$.

Likewise $Ann(ann(N))=N$ if and only if $N=Ann(L)$ for some left ideal $L$ of $R$.

When $M$ is the ring itself, ideals like these are called "annihilator ideals" or sometimes "annulets" but I am not completely sure what they call them in the general case when looking at the annihilator maps between a module and its ring.