Let us identify each card with the set of digits written on it. For any collection of cards $C_{1}, C_{2}, \ldots, C_{k}$ denote by their sum the set $C_{1} \triangle C_{2} \triangle \cdots \triangle C_{k}$ consisting of all elements belonging to an odd number of the $C_{i}$'s. Denote the first and the second player by $\mathcal{F}$ and $\mathcal{S}$, respectively.

Since each digit is written on exactly 512 cards, the sum of all the cards is $\varnothing$. Therefore, at the end of the game the sum of all the cards of $\mathcal{F}$ will be the same as that of $\mathcal{S}$; denote this sum by $C$. Then the player who took $C$ can throw it out and get the desired situation, while the other one cannot. Thus, the player getting card $C$ wins, and no draw is possible.

Now, given a nonempty card $B$, one can easily see that all the cards can be split into 512 pairs of the form ($X$, $X \triangle B$) because ($X \triangle B$) $\triangle B = X$. The following lemma shows a property of such a partition that is important for the solution.

Lemma. Let $B \neq \varnothing$ be some card. Let us choose 512 cards so that exactly one card is chosen from every pair $(X, X \triangle B)$. Then the sum of all chosen cards is either $\varnothing$ or $B$.

Proof. Let $b$ be some element of $B$. Enumerate the pairs; let $X_{i}$ be the card not containing $b$ in the $i^{\text{th}}$ pair, and let $Y_{i}$ be the other card in this pair. Then the sets $X_{i}$ are exactly all the sets not containing $b$, therefore each digit $a \neq b$ is written on exactly 256 of these cards, so $X_{1} \triangle X_{2} \triangle \cdots \triangle X_{512} = \varnothing$. Now, if we replace some summands in this sum by the other elements from their pairs, we will simply add $B$ several times to this sum, thus the sum will either remain unchanged or change by $B$, as required. $\square$

Now we consider two cases.

Case 1. Assume that $\mathcal{F}$ takes the card $\varnothing$ on his first move. In this case, we present a winning strategy for $\mathcal{S}$.

Let $\mathcal{S}$ take an arbitrary card $A$. Assume that $\mathcal{F}$ takes card $B$ after that; then $\mathcal{S}$ takes $A \triangle B$. Split all 1024 cards into 512 pairs of the form $(X, X \triangle B)$; we call two cards in one pair partners. Then the four cards taken so far form two pairs ($\varnothing, B$) and ($A, A \triangle B$) belonging to $\mathcal{F}$ and $\mathcal{S}$, respectively. On each of the subsequent moves, when $\mathcal{F}$ takes some card, $\mathcal{S}$ should take the partner of this card in response.

Consider the situation at the end of the game. Let us for a moment replace card $A$ belonging to $\mathcal{S}$ by $\varnothing$. Then he would have one card from each pair; by our lemma, the sum of all these cards would be either $\varnothing$ or $B$. Now, replacing $\varnothing$ back by $A$ we get that the actual sum of the cards of $\mathcal{S}$ is either $A$ or $A \triangle B$, and he has both these cards. Thus $\mathcal{S}$ wins.

Case 2. Now assume that $\mathcal{F}$ takes some card $A \neq \varnothing$ on his first move. Let us present a winning strategy for $\mathcal{F}$ in this case.

Assume that $\mathcal{S}$ takes some card $B \neq \varnothing$ on his first move; then $\mathcal{F}$ takes $A \triangle B$. Again, let us split all the cards into pairs of the form $(X, X \triangle B)$; then the cards which have not been taken yet form several complete pairs and one extra element (card $\varnothing$ has not been taken while its partner $B$ has). Now, on each of the subsequent moves, if $\mathcal{S}$ takes some element from a complete pair, then $\mathcal{F}$ takes its partner. If $\mathcal{S}$ takes the extra element, then $\mathcal{F}$ takes an arbitrary card $Y$, and the partner of $Y$ becomes the new extra element.

Thus, on his last move $\mathcal{S}$ is forced to take the extra element. After that player $\mathcal{F}$ has cards $A$ and $A \triangle B$, player $\mathcal{S}$ has cards $B$ and $\varnothing$, and $\mathcal{F}$ has exactly one element from every other pair. Thus the situation is the same as in the previous case with roles reversed, and $\mathcal{F}$ wins.

Finally, if $\mathcal{S}$ takes $\varnothing$ on his first move then $\mathcal{F}$ denotes any card which has not been taken yet by $B$ and takes $A \triangle B$. After that, the same strategy as above is applicable.