Solution:

(a) If $C_{1}$ is free, $T_{1}$ can be moved to the left or placed on $C_{1}$, depending on whether it is already on the grid or not; the same reasoning applies to $C_{2018}$, so from now on we will assume that $C_{1}$ and $C_{2018}$ are both occupied. Having chosen a free cell $C_{k}$, let us call $T_{a}$ and $T_{b}$ the first tiles encountered starting from $C_{k}$ and proceeding respectively to the left and to the right (hence $a < b$): if $k \leq a$ one can move $T_{a}$ to the right, just as if $k \geq b$ one can move $T_{b}$ to the left; finally, if $a < k < b$, one can place $T_{k}$ on $C_{k}$.

(b) We prove by induction on $N$ that, playing on a $1 \times N$ grid, Alice has a winning strategy whose first move consists of placing $T_{1}$ on $C_{N}$ (or equivalently, if there is only $T_{1}$ on $C_{N}$ and it is Barbara's turn, Alice has a winning strategy).
If $N=2$, Barbara must necessarily move $T_{1}$ to $C_{1}$, after which Alice wins by placing $T_{2}$ on $C_{2}$.
Now suppose that the claim holds for some $N \geq 2$ and that play takes place on a $1 \times (N+1)$ grid: as before, Barbara must move $T_{1}$ to $C_{N}$, and Alice can respond by placing $T_{N+1}$ on $C_{N+1}$: since this last tile can no longer be moved, the rest of the game will take place on the $1 \times N$ grid obtained by excluding $C_{N+1}$ (and one easily observes that the presence or absence of $C_{N+1}$ does not affect the set of legal moves). On this grid, after the moves described, there is only $T_{1}$, which is on $C_{N}$, and it is Barbara's turn, but by the induction hypothesis this configuration is favorable to Alice.