Finding Language of a CFG
Let's consider "half" of your problem: $S\to AbASc \mid \epsilon$, and $A\to aA \mid \epsilon$. It is important to realize that the number of $b$'s and $c$'s generated is equal. Try to express that in your solution. 

At the same time, when two $A$'s are generated, they will generate an arbitrary number of $a$'s. One of your solutions claims these numbers are always the same, but they are not.

Personally I would prefer to write this language as $L(G_{\frac12}) = \{ wc^k \mid w\in\{a,b\}^*, k\ge1, \#_b(w) = k\} \cup \{\epsilon\}$, where $\#_b(w)$ is the notation for the number of occurrences of letter $b$ in the string $w$.
Alternatively, if you want to avoid that notation perhaps $L(G_{\frac12}) = \{ a^{n_0}ba^{n_1}b\dots ba^{n_k}c^k \mid k\ge1, n_0,\dots,n_k\ge 0\}\cup \{\epsilon\}$.

It is not immediate how to write this using "regular expressions" as the equality of the number of $b$'s and $c$'s is not regular property itself. Also we cannot put the star-operator inside set brackets, as the star yields a language itself, so it would be proper to write
$L(G_{\frac12}) = \bigcup_{k\ge1} (a^*b)^ka^*c^k \cup \{\epsilon\} = \bigcup_{k\ge0} (a^*ba^*)^kc^k $ .

Sorry about the $\dots \cup \{\epsilon\}$ in several of these solutions. Allowing $k=0$ would not give the right answer, as that would add $a^*$ to the language, which is not generated by the grammar.