(a) For $n = 2015$, you can make all cards white-side-up.
Label the cards from $1$ to $2015$ in order. WLOG assume the only black card initially is $2$. Consider the following moves (B=black, W=white):
* $1$B, $2$W, $3$B
* $1$B, $2$B, $3$W, $4$B
* $1$B, $2$B, $3$B, $4$W, $5$B
* ...
* $1$B, $2$B, ..., $2013$B, $2014$W, $2015$B
* $1$W, $2$B, $3$B, ..., $2014$B, $2015$W.
Now, there are $2013$ consecutive black-side-up cards. Group them by three, and flip them over in groups (which takes $2013/3 = 671$ moves), and we get all cards white-side-up.

(b) For $n = 2016$, you cannot make all cards white-side-up.
Label the card as $\alpha, \beta, \gamma, \alpha, \beta, \gamma, \dots$ in order; note that there are equal numbers of $\alpha, \beta$ and $\gamma$. Now, for each move, you always flip an $\alpha$, an $\beta$, and a $\gamma$. Suppose the initial black-side-up card is labelled $\alpha$. Then, you need an odd number of moves to let all $\alpha$ white-side-up, but you need an even number of moves to let all $\beta$ (or $\gamma$) card white-side-up. So it is impossible.