a) If the game does not end, then, since there are at most $2^{2008}$ different possible game states, there must exist a periodically repeating sequence of states. Let $k$ be the smallest number of the cards that are turned over within this period. However, to turn the card with number $k$ from black to white, a card with a number smaller than $k$ must be chosen – a contradiction to minimality! Therefore, the game must end after finitely many moves.

b) We consider the cards with numbers $41k$, where $1 \leq k \leq 48$. The highest of these cards, with number $1968$, is the last one that can be chosen. With each move, exactly one of these cards is turned over. Since the game ends when all cards up to number $1968$ are black, and since these cards are all white at the start, after every double move of $A$ and $B$ an even number of them is white again; thus $B$ always finds an odd number of white cards from this set and can therefore always move. Hence $B$ inevitably wins the game, no matter how he plays.