Alice wins for $n$ odd and Bob wins for $n$ even. Roughly speaking, when $n$ is odd Alice colours any vertex and then Alice imagines a line going through that vertex that separates the polygon into two symmetric halves. From now on, Alice simply colours the reflection of the vertex coloured by Bob. This assures that Alice gains at least as many points as Bob on each subsequent move(note that Alice did not gain any points in the first move). This could go down to the wire but Bob eventually loses when he colours one of the points that are simultaneously neighbours and reflections of each other since then Alice will reap the extra benefit of the point coloured by Bob. When $n$ is even, Bob sees the first vertex coloured by Alice, randomly selects one of its incident edges and imagines a line going through the midpoint of that edge splitting the polygon in two symmetric halves. He then simply colours the reflection with respect to that line of whichever vertex Alice coloured to ensure that he does at least as well as Alice on each move. In this case, the draw does not even have a chance of happening because in his first move, Bob colours a vertex whose neighbour is already coloured by Alice thus Bob wins in this case.

Good problem for beginners in games. Alice wins for odd $n$, Bob wins for even $n$. For even $n$, after Alice plays arbitrarily, say on vertex $V$ (and gets $0$), Bob colours a vertex $W$ to the one already coloured by Alice (and gets one point); after that whatever Alice colours, Bob colours the point symmetric to Alice's last one with respect to the perpendicular bisector of $VW$ -- in this manner on each turn Bob gets at least at many points as Alice's last turn. Because of their first moves, Bob surely has more points overall. For odd $n$, Alice firstly plays arbitrarily, say on vertex $V$ (and gets $0$), Bob colours a vertex $W$ in a way he wants and then Alice picks the unique vertex $Z$ such that the shortest distance (on the $n$-gon) between $V$ and $W$ is the same at the shortest between $W$ and $Z$. (Note that both players get $1$ if $W$ is next to $V$ and get $0$ otherwise.) Now there are an even amount of remaining vertices, it is Bob's turn and from now on, whatever he colours, Alice colours the point symmetric to Bob's last one with respect to the perpendicular bisector of $VZ$ (note that such a point always exists, as the only vertex of the $n$-gon which is its own reflection is $W$, but it is already coloured). In that manner, Alice gets at least as many points as Bob does. To justify that these amounts are not equal, it suffices note that the total number of points for both players is equal to $n$. Indeed, put an arrow from vertex $X$ to vertex $Y$ if, whenever we colour $X$, we get a point from $Y$ because it turns out to be an already coloured neighbour; then clearly there is exactly one arrow on each side of the $n$-gon and no arrows elsewhere. EDIT: Ok, I now see that there are even simpler ways to write down the play when $n$ is odd, whatever.

Notice the number of points Alice and Bob get is equal to the number of polygon sides each of them complete. So, the number of points corresponds to the number of sides in an $n$-sided polygon, which is $n$. Now, I claim that for even $n$, i.e. $n=2k$, Bob has a winning strategy, and for odd $n$, i.e. $n=2k+1$, Alice has a winning strategy. First notice that Alice, being the first person to move, gets 0 points, and likewise, the last person to move gets 2 points. Next, notice that on every subsequent turn, it is always possible for the player to get at least 1 point, by colouring in an uncoloured point adjacent to a coloured point. If it doesn't exist, it can mean one of two things: Alice hasn't moved, so every vertex is uncoloured, a contradiction; or that every vertex has been coloured, which happens after $n$ turns have been performed. So, for even $n=2k$ Alice and Bob each move $k$ times, with Alice's first move earning her 0 moves, and Bob's last move earning him 2 points. Hence Bob can always get at least $(k-1) + 2 = k+1$ points, while Alice can get at most $n-(k+1) = k-1$ points, Bob wins. Otherwise, for odd $n=2k+1$, Alice moves first and last. Ignoring her first move, Alice and Bob each have $k$ moves, and Alice can always ensure she gets 1 point on every turn, and gets 2 points in her last turn, a minimum of $(k-1) + 2 = k+1$ points. Likewise, Bob can only get at most $k$ points, so Alice wins.

Same solution as above, this problem is very trivial lol Bob wins for even $n$, Alice wins for odd $n$. Lemma: the number of points that Bob and Alice win in total, is the number of vertices. Proof: trivial. Just draw a graph with the points and each time you draw a line segment connecting two vertices, it represents a point. The total number of line segments is the number of vertices. If $n = 2k+1 \hspace{0.20cm} k \in \mathbb{N}$, then Alice will win. There are $2k+1$ points to be won in total. Notice that Alice begins with a move and also plays the last move. In her last move, she will always win $2$ points. Therefore, in the $2k-1$ moves between first and last move, Alice makes $k-1$ moves and Bob makes $k$ moves. Notice that Alice can always guarantee herself one point by colouring a vertex next to an already coloured vertex. Therefore, she will gain atleast $(k-1)+2 = k+1$ points, making herself the winner. If $n = 2k \hspace{0.20cm} k \in \mathbb{N}$, then Bob will win. He gets $2$ points from the last move and can always get $1$ point in the other $k-1$ moves he has, meaning he gets $k+1$ points atleast, making him the winner. Is it that simple?