Let $n$ be the number of different questions Asma asks to know the values of each number $a_{1}, a_{2}, \ldots, a_{2013}$. Let us make the following remarks:

a. Because Asma needs to know the value of each number $a_{i}$, each number $a_{i}$ must appear at least in one of the sets that Asma chooses to ask about.

b. If Asma chooses a set $\{a_{i_{1}}, a_{i_{2}}\}$, she needs to choose another set that contains only one of the numbers $a_{i_{1}}, a_{i_{2}}$, so that she can distinguish between the values of $a_{i_{1}}$ and $a_{i_{2}}$.

c. If Asma asks about the sets $\{a_{i_{1}}, a_{i_{2}}\}$ and $\{a_{i_{2}}, a_{i_{3}}\}$, for different indices $i_{1}, i_{2}, i_{3}$, she will deduce the value of each number $a_{i_{1}}, a_{i_{2}}, a_{i_{3}}$, by taking the intersection of the two sets and their differences.

d. If for some $k \geq 3$, Asma asks about the sets $\{a_{i_{1}}, a_{i_{2}}\}, \{a_{i_{2}}, a_{i_{3}}\}, \ldots, \{a_{i_{k-1}}, a_{i_{k}}\}$, for different indices $i_{1}, i_{2}, \ldots, i_{k}$, she will deduce the value of each number $a_{i_{1}}, a_{i_{2}}, \ldots, a_{i_{k}}$ and does not need to ask after this about the set $\{a_{i_{k}}, a_{i_{1}}\}$ since she already knows the answer.

From now on, we assume, without loss of generality, that Asma never asks about a set that she already knows.

According to these remarks, we can make a partition of the set $\{a_{1}, a_{2}, \ldots, a_{2013}\}$ into non-empty disjoint subsets $A_{1}, A_{2}, \ldots, A_{m}$ such that two numbers $a_{i}, a_{j}$ are in the same subset $A_{k}$ if and only if there exist sets $\{a_{i_{1}}, a_{i_{2}}\}, \{a_{i_{2}}, a_{i_{3}}\}, \ldots, \{a_{i_{l-1}}, a_{i_{l}}\}$ that Asma has asked about each one of them, and such that $a_{i_{1}} = a_{i}, a_{i_{l}} = a_{j}$. Notice that if $a_{i_{1}}, a_{i_{2}}, \ldots, a_{i_{l}}$ are different, because of remark (d) they are unique.

For each $A_{i}$, there are exactly $\#A_{i} - 1$ sets in $A_{i}$ that Asma has asked about. To see this, fix $a_{i_{0}} \in A_{i}$, and for each $a_{j} \in A_{i}$ with $a_{j} \neq a_{i_{0}}$, associate the unique set $\{a_{i_{l-1}}, a_{i_{l}}\}$ in the previous paragraph when $a_{i_{1}} = a_{i_{0}}, a_{i_{l}} = a_{j}$ and $a_{i_{1}}, a_{i_{2}}, \ldots, a_{i_{l}}$ are different. It is not hard to see from remark (d) that this map is a bijection between $A_{i} \setminus \{a_{i_{0}}\}$ and the set of subsets of $A_{i}$ in the questions of Asma.

Because each set in the questions of Asma is contained exactly in one of the subsets $A_{1}, A_{2}, \ldots, A_{m}$, we have
$$n = \sum_{i=1}^{m} (\#A_{i} - 1) = 2013 - m.$$
Because each set $A_{i}$ must contain at least $3$ elements (remarks (a) and (b)),
$$m \leq \left\lfloor \frac{2013}{3} \right\rfloor = 671$$
This proves that $n \geq 1342$.

Conversely, if Asma asks about the $1342$ sets
$$\{a_{3i+1}, a_{3i+2}\}, \{a_{3i+2}, a_{3i+3}\}, \quad \text{for} \quad i = 0, 1, \ldots, 670$$
she will know the values of all the numbers $a_{1}, a_{2}, \ldots, a_{2013}$.