Context free grammar for a palindrome of even length + cc
This CFG does not work
$$\begin{aligned}
S&amp;\to aS \mid bS\mid\epsilon\mid T\\
T&amp;\to  cc\\
\end{aligned}$$
It generates $acc$ by $S\Rightarrow aS\Rightarrow aT\Rightarrow acc$, which is not of form $AA^Rcc$. In fact, it generates the language $(a+b)^*(\epsilon+cc)$.

You might have missed the basic mindset to write a context-free grammar (CFG).
Every non-terminal in a CFG stands for a language.
Thinking in terms of languages instead of strings should be more helpful sometimes for constructing the CFG.
Since the language $S=\{AA^Rcc: A \in \{a,b\}^*\}$ is the concatenation of the language of the palindromes of even length,  $P=\{AA^R: A \in \{a,b\}^*\}$ and the language of one word, $C=\{cc\}$, the corresponding substitution rule is
$$ S\to PC$$
to which we add the substitution rules for $P$ and $C$
$$\begin{aligned}
P&amp;\to aPa\mid bPb\mid\epsilon\\
C&amp;\to  cc.\\
\end{aligned}$$
The collection of all substitution rules above is the CFG wanted. Note that each of the symbols $S, P, C$ is abused to denote a non-terminal as well as the languages it stands for, i.e, the context-free language generated if we treat it as the starting symbol.
Exercise
Write a CFG for the language $\{AA^RbcBB^R: A \in \{a,b\}^*, B\in\{c,d\}^*\}$.