Suppose the coins are $M_1, M_2, ..., M_{10}$, and for each $n=1,2,...,9$ let $X_n = \{M_n, M_{n+1}\}$. We know one of the $X_n$ contains the two fake coins. We'll prove that it is possible to reach the goal with two questions. To see that, suppose $S$ is a set of coins. When we ask how many fake coins are in $S$, the answer we are given is the number of elements there are in the intersection between $S$ and the $X_n$ which contains the fake coins. Hence, it is enough to find two subsets $S_1, S_2$ of $\{M_1, M_2, ..., M_{10}\}$ such that the answers given in the two questions are different for each $X_n$. Take $S_1 = \{M_4, M_7, M_8, M_9, M_{10}\}$, $S_2 = \{M_1, M_5, M_6, M_7, M_8\}$. If the answers given are $(0,0) \Rightarrow M_2, M_3$ are the fake coins $(0,1) \Rightarrow M_1, M_2$ are the fake coins $(0,2) \Rightarrow M_5, M_6$ are the fake coins $(1,0) \Rightarrow M_3, M_4$ are the fake coins $(1,1) \Rightarrow M_4, M_5$ are the fake coins $(1,2) \Rightarrow M_6, M_7$ are the fake coins $(2,0) \Rightarrow M_9, M_{10}$ are the fake coins $(2,1) \Rightarrow M_8, M_9$ are the fake coins $(2,2) \Rightarrow M_7, M_8$ are the fake coins The solution is complete

Let $c_1,c_2,...,c_{10}$ the coins, because the two fake coins are consecutive, there are $9$ posible positions for these fake coins (namely, $F=\{b_1,b_2,...,b_9\}$, where $b_1=c_1c_2, b_2=c_2c_3,...,b_9=c_9c_{10}$). Supose that it is posible to know the positions of the two fake coins. We fix two subsets $A,B$ of $\{c_1,c_2,...,c_{10}\}$, and then we get a vector $(a,b)$ in $\mathbb{Z}_{3}\times \mathbb{Z}_{3}$, were the first coordinate $a$ is the number of fake coins in the set $A$, and the second coordinate $b$ is the number of fake coins in the set $B$ (we consider $\mathbb{Z}_3=\{0,1,2\}$ cause each set $A,B$ can have $0,1$ or $2$ fake coins). Because there are $|\mathbb{Z}_{3}\times \mathbb{Z}_{3}|=3\cdot 3=9$ posible vectors, we define a function $f: F\to \mathbb{Z}_{3}\times \mathbb{Z}_{3}$, where $f(b_i)=(x,y)$ if $b_i$ have $x$ coins in $A$ and $y$ coins in $B$ ($1\le i\le 9$). We claim that $f$ is injective. Assume that $f(b_i)=f(b_j)$ if $i\not= j$, then if the two consecutive fake coins are in $b_i$, then we can chose $b_j$ as the set with fake coins and we cannot determinate the positions of the fake coins. Then $f$ is injective and because $|F|=|\mathbb{Z}_{3}\times \mathbb{Z}_{3}|$, $f$ is indeed bijective. Now the work is easy and it's not difficult to see that $A=\{c_3,c_4,c_8,c_9,c_{10}\}$ and $B=\{c_3,c_4,c_5,c_6,c_{10}\}$ satisfies.

The answer is positive. Number the coins $1$ through $10$ in order from left to right. In the first question, we ask about coins $1, 3, 4, 5, 6$ and for the second question we ask about coins $2, 3, 4, 9, 10$. Then, we can determine which coins are false according to the following table: \begin{tabular}{|c|c|c|} \hline Answer to Q1 & Answer to Q2 & False Coins \\ \hline 0 & 0 & 7, 8 \\ 0 & 1 & 8, 9 \\ 0 & 2 & 9, 10 \\ 1 & 0 & 6, 7 \\ 1 & 1 & 1, 2 \\ 1 & 2 & 2, 3 \\ 2 & 0 & 5, 6 \\ 2 & 1 & 4, 5 \\ 2 & 2 & 3, 4 \\ \hline \end{tabular}This proves it is possible. $\blacksquare$

If we use the sets $\{3, 4, 8, 9, 10\}$ and $\{3, 4, 5, 6, 10\}$, it satisfies the condition. Shamelessly copying @above's table we get \begin{tabular}{|c|c|c|} \hline Answer to Q1 & Answer to Q2 & False Coins \\ \hline 0 & 0 & (1, 2) \\ 0 & 1 & (6, 7) \\ 0 & 2 & (5, 6) \\ 1 & 0 & (7, 8) \\ 1 & 1 & (2, 3) \\ 1 & 2 & (4, 5) \\ 2 & 0 & (8, 9) \\ 2 & 1 & (9, 10) \\ 2 & 2 & (3, 4) \\ \hline \end{tabular}