Decidability of "Is this regular expression prefix-free?"
Let $L$ be a language on the alphabet $A$. Then $L$ is prefix-free if $L \cap LA^+$ is empty. If $L$ is regular and given by a regular expression, then $L \cap LA^+$ is also regular (and one can effectively compute a regular expression for it). Now testing whether a given regular language is empty is decidable.
For context-free languages, the situation is different. Above proof does not work because the context-free languages are not closed against intersection. In fact, there is no algorithm to decide whether a given context-free grammar generates a prefix-free language.
Here is a proof of this claim [Luc Boasson, personal communication].
Let $(u_1, v_1), \dotsm, (u_n,v_n)$ be two finite lists of words of $A^*$ defining an instance of the Post Correspondence Problem (PCP) and let $X = \{x_1, \ldots, x_n\}$ be an alphabet of $n$ letters. Assume that $A$ and $X$ are disjoint and consider the homomorphims $u, v: X^* \to A^*$ defined by $u(x_i) = u_i$ and $v(x_i) = v_i$. Consider the languages
\begin{align}
 U &amp;= \{x^Ru(x) \mid x \in X^+ \} \\
 V &amp;= \{x^Rv(x) \mid x \in X^+ \} 
\end{align}
Let $\sharp$ be a new letter. Then $U \cup V\sharp$ is a context-free language. It is prefix-free if and only if $U$ and $V$ are disjoint if and only if the Post Correspondence Problem has no solution. Thus deciding whether a given context-free language is prefix-free is undecidable.