The pair $(X,Y)$ is not bivariate normal since $X+Y$ has positive probability of being equal to $0$, but is not always equal to $0$, so it's not normally distributed. (It's equal to $0$ whenever $|X|>a$. Besides, just look at the graph of $Y$ as a function of $X$ and you see that it's constrained to lie on an odd-shaped one-dimensional figure that is not a straight line.
Notice that $f_X(-y)$ is the same as $f_X(y)$ because of the symmetry of the distribution of $X$. That is indeed the density function of $Y$, and that tells you that the distribution of $Y$ is $N(0,1)$. So they're separately normal but not jointly normal.
The covariance depends on $a$ in a way that becomes clearer if you think about what happens to the correlation when $a$ is very close to $0$ and when $a$ is very very big. You see that in one case you get positive correlation and in the other you get negative correlation, so somewhere in between you get $0$ correlation. This shows that a pair of random variables can be separately normally distributied with covariance $0$ without being independent. (If they were jointly normally distributed and uncorrelated, they'd be independent.)