Replace $2013$ by a general odd number $2k-1$, $k \ge 2$. The sum $S = 1+2+\dots+(2k-1)$ equals $k(2k-1)$. For part a), the only number that can be determined with certainty is $k$, the one in the middle. To find $k$, ignore a card with an unknown number $x$ on it, thus forming a set of $2k-2$ cards, and ask about their average. It is an integer if and only if $2k-2$ divides $S-x = k(2k-1) - x = k(2k-2) + (k-x)$, i.e. if and only if $2k-2$ divides $k-x$. Now $1 \le x \le 2k-1$ gives $|k-x| \le k-1 < 2k-2$ (by $k \ge 2$), thus $x = k$ is the only possibility. Hence applying the procedure to each card will exhibit $k$.

Conversely, suppose that the card with a number $m$ can be determined with certainty by asking a sequence of questions. Imagine that the number $j$ on each card is replaced by $2k-j$; the numbers $1, 2, \ldots, 2k-1$ are written on the cards again. Now ask the same sequence of questions. Since the average of $2k-j, \ldots, 2k-j$ is integer if and only if the average of $j_1, \ldots, j_n$ is, the answers will be the same as with the initial situation. Hence the questions that initially find the card with $m$ will find the card with $2k-m$ in the new situation; so $2k-m = m$ and $m = k$. This completes a).

b) Let the cards be divided into groups so that the contents of each group as a whole is known. Since $k$ is the unique number that can be found with certainty, all groups contain at least $2$ cards each except possibly one, which contains the card with $k$ and has size $1$. It follows that the number of groups is at most $\frac{1}{2} ((2k-1)-1)+1 = k$. We prove that $k$ groups can be obtained. More precisely, excluding the card with $k$, the remaining cards can be divided into $k-1$ pairs such that each pair contains numbers of the form $j, 2k-j$, $j=1, \ldots, k-1$. Call such cards complementary.

Note that, once $k$ is found, one can determine the parity of the number on each card. Let e.g. $k$ be odd. Choose any card $x$ and ask about the two cards $k, x$. If their average is an integer then $x$ is odd, otherwise is even. The case of even $k$ is analogous.

Let us show now how to find the complementary pairs $C_j = \{j, 2k-j\}$, $j=1, \ldots, k-1$. Start with $C_1 = \{1, 2k-1\}$ and $C_2 = \{2, 2k-2\}$. Take a pair of cards with different parity and unknown sum $y$; $y$ is odd. Ask about the average of the remaining $2k-3$ cards. It is an integer if and only if $2k-3$ divides $S-y = k(2k-1)-y = k(2k-3) + 2k - y$, hence if and only if $2k-3$ divides $2k-y$. Note that $3 \le y \le 4k-3$, yielding $|2k-y| \le 2k-3$. So the answer is yes if and only if $2k-y \in \{0, \pm(2k-3)\}$. Because $2k-y$ is odd (as $y$ is), we obtain $2k-y = \pm(2k-3)$, yielding $y=3$ or $y=4k-3$. These are the extremal values of the sum $y$; they are achieved only if the numbers in the pair are $1, 2$ or $2k-2, 2k-1$ respectively.

By repeating this procedure with all pairs of cards with different parity we find the two pairs $1, 2$ and $2k-2, 2k-1$ (without knowing which one is which). Thus a group containing the $4$ cards $1, 2, 2k-2, 2k-1$ is determined. It has $2$ odd and $2$ even numbers, and the parity of the number on each card is known. Hence the two odd cards in the group form the complementary pair $C_1 = \{1, 2k-1\}$, the two even cards form $C_2 = \{2, 2k-2\}$.

Suppose that the complementary pairs $C_1, \ldots, C_{2j}$ are determined for some $j$ such that $2j \le k-1$. We show how to find $C_{2j+1}$ and $C_{2j+2}$. One may assume $2j \le k-3$. Indeed if $2j = k-1$ (with $k$ odd) then all complementary pairs are already found. If $2j = k-2$ (with $k$ even) then there is only $1$ complementary pair left, $C_{k-1}$. But once $C_{k-1}, \ldots, C_{k-2}$ are known, so is $C_{k-1}$.

Exclude the $4j$ numbers $C_1, \ldots, C_{2j}$. There remain $2k-4j-1$ numbers with sum $S-4jk$. Again take a pair of cards with different parity and unknown odd sum $y$; note that $4j+3 \le y \le 4k-4j-3$. Like before ask about the average of the remaining $2k-4j-3$ cards. It is an integer if and only if $2k-4j-3$ divides $(S-4jk)-y = k(2k-1)-4jk-y = k(2k-4j-3)+(2k-y)$, i.e., if and only if $2k-4j-3$ divides $2k-y$. Now $4j+3 \le y \le 4k-4j-3$ gives $|2k-y| \le 2k-4j-3$. So the answer is yes if and only if $2k-y \in \{0, \pm(2k-4j-3)\}$. Because $2k-y$ is odd, we obtain $2k-y = \pm(2k-4j-3)$, yielding $y=4j+3$ or $y=4k-4j-3$. These are the extremal values of $y$, achieved only if the numbers in the pair are $2j+1, 2j+2$ or $2k-2j-2, 2k-2j-1$ respectively.

Hence the procedure applied to all pairs of the kind considered, yields a group containing the $4$ cards $2j+1, 2j+2, 2k-2j-2, 2k-2j-1$. The two odd cards in the group form the complementary pair $C_{2j+1} = \{2j+1, 2k-2j-1\}$, the two even ones form $C_{2j+2} = \{2j+2, 2k-2j-2\}$.

We presented an inductive argument which divides all cards different from $k$ into complementary pairs. This completes the solution.