On a paper, you create a row of $n$ ones. In one move, Art or Bart, choses two ones next to each other, placing either a $+$ or $\times$, in between the ones. After $n-1$ moves there will be either a $+$ or $\times$ between every pair of ones and the game is over. Then you calculate the total number $T$ containing all these ones: If $T$ is even, Art wins. If $T$ is odd, Bart wins.
Does either Art or Bart have a winning strategy? (Art starts of the game).
I assume that multiplication will be evaluated before addition (using your preferred acronym for order of operations), and hence the result of any product will be $1$, as the product of $k$ $1$s for any $k$ is still inevitably $1$.
We can then see that $T$ simplifies to be a string of $1$s added together, with $n-k-1$ addition signs, hence the result will be $n-k$, where $k$ is the total number of multiplication signs used. Once we take care of the trivial case for $n=1$ where Bart wins immediately, we can see that whoever plays the last move can alter the parity to win.
We can see that if $n$ is odd then Bart plays last and wins, and if $n$ is even then Art plays last and wins. So yes a winning strategy does exist for Art or Bart depending on $n$, but there is no universal winning strategy for either for all $n$.