Saturation and Jacobson radical

Both of these are garbled:

I think $S_{A'} = ${$a' \in A' | \exists b \in A' $ such that $a'b\in S_{A'} $}.

So, Saturated set is all $a'\in S_{A'}$ such that $ \exists b\in S_{A'}$ such that $a'b =1+a''$[, where] $a'' \in S_{A'}$ .

You miswrote your definition two times in a row. It should read

I think $\bar S_{A'} = \{a \in A | \exists b \in A \text{ such that }ab\in S_{A'} \}$.

and

So, Saturated set is all $a\in A$ such that $ \exists b\in A$ such that $ab =1+a'$[, where] $a' \in A'$ .

Given that latter one, you have $ab-1\in A'$, and I don't see any other better explanation than

$\bar S_{A'}$ is the set of all $a\in A$ such that $a+A'$ is a unit in $A/A'$.

It's funny this should happen but I'm now remembering an anecdote by Arhangelskii where he recounted a mathematician with the unfortunate habit of using $A$'s for everything: $a$, $A$, $\alpha$, $\mathfrak A$, $\mathscr A$ with various diacritics or subscripts.

Next time, use $R$ and $I\lhd R$ please instead of $A'\lhd A$.

For $c$, it suffices ot use the characterization of the Jacobson radical as the set of all elements $x$ such that $xr-1$ is a unit for every $r$ in the ring.

To that end, look at something of that form: $\frac{i}{j+1}\frac{r}{k+1}-1$ where $i, j, k\in I$ and $r\in R$.

If you write $\frac{i}{j+1}\frac{r}{k+1}=\frac{ir}{m+1}$ with $m\in I$, then $\frac{ir}{m+1}-1=\frac{ir-m-1}{m+1}$ is a unit iff $ir-m-1$ is a unit.

You should be able to take it from there...