Undecidability of "is this CFG prefix-free?"
One endmarker is insufficient to prove this. Instead, we must use multiple endmarkers. Here's the full proof, with the same basic idea taken from the other answer.
Using the notation from Sipser 3rd edition problem 5.21, let the string pairs $(t_i, b_i)$ for $i \in \{1, 2 \ldots k\}$ be an instance of the PCP problem. Define grammar rules $T \rightarrow t_i T \texttt{a}_i$ and $T \rightarrow t_i \texttt{a}_i$ (where $\texttt{a}_i$ are new symbols) along with $B \rightarrow b_i B \texttt{a}_i$ and $B \rightarrow b_i \texttt{a}_i$. This instance of PCP has a solution if and only if $T \cap B$ is non-empty. (This construction can be used to prove many other undecidability properties of CFGs; for more, see Sipser 3rd edition chapter 5 or Hopcroft and Ullman 3rd edition chapter 9).
Define more rules $W \rightarrow T\#$ and $W \rightarrow B\#\#$, where $\#$ is another new symbol. Clearly, $W$ is prefix free if and only if $T \cap B$ is non-empty. Hence, a decider for prefix-free CFGs would also decide any PCP problem. Since PCP is undecidable, we have a contradiction, and thus the language of prefix free CFGs is also undecidable.