Considering a grammar that includes a dictionary or list with an optional terminal separator - e.g.: $$ \begin{align} obj \to    & dict  \\              &\mid OTHER  \\ dict \to   & \{ \quad \}  \\            & \mid \{ \quad pairs \quad \} \\ pairs \to  & pair \quad ,?  \\           & \mid pair \quad , \quad pairs  \\ pair \to   & PAIR\end{align}$$My understanding is that the above is not an LL(1) grammar because there are productions with FIRST() setscrossing over.But what if we convert it to the following:$$ \begin{align} obj \to    & dict  \\              &\mid OTHER  \\ dict \to     & \{ \quad dictrest \\dictrest \to & \}  \\                 & \mid pair \quad pairseq \quad ,? \quad \}  \\pairseq \to  & \varepsilon \\             & \mid , \quad pair \quad pairseq \\pair \to   & PAIR\end{align}$$Is this latter version LL(1)?
Representing grammar for list/dictionary with optional end separator as LL(1)
formal grammars;parsers
You grammar can be boiled down by removing some chain rules to this:$\qquad\begin{align} obj &\to \{\ dict \mid OTHER \\ dict &\to\ \} \mid PAIR\ pairs\ ,\ \} \mid PAIR\ pairs\ \} \\ pairs &\to\ ,\ PAIR\ pairs \mid \varepsilon\end{align}$Note that I expanded the $?$ which, as far as I know, is not EBNF. This introduces one source of ambiguity LL(1) can not get rid of. The version with $?$, i.e.$\qquad\begin{align} obj &\to \{\ dict \mid OTHER \\ dict &\to\ \} \mid PAIR\ pairs\ ,?\ \} \\ pairs &\to\ ,\ PAIR\ pairs \mid \varepsilon\end{align}$has a similar problem: while deriving $dict$ and you see $PAIR\ ,$, which alternative of $pairs$ do you choose? Impossible to decide.Note that your basic problem is that you create a sequence of stuff from the left without knowing what the last element is going to be. Formally, you have the right-linear grammar$\qquad A \to abA \mid ab \mid a$which generates $(ab)^*a(b\mid\varepsilon)$, surrounded by$\qquad S \to \{\ A\ \}$.Selecting the very last rule is indeed problematic -- because you threw away the real end marker, the closing brace! Move that one to $A$, get rid of the ambiguity with $a$ and you are good! $\qquad\begin{align} S &\to \{\ A \\ A &\to aB \mid\ \} \\ B &\to bA \mid \ \}\end{align}$