Finding the 2 fake balls. We have $three$ red $\{a_1,a_2,a_3\}$ and $five$, $\{b_1,b_2,b_3,b_4,b_5\}$ green balls. 
Two of them are "fake", one in red balls and the other in green balls. We can choose arbitrary set of these balls and ask if this set contains a fake ball.  We need to find both the red fake ball and the green fake ball after $4$ such questions. 
I can do this using $5$ questions, and testing red and green separately` $2$ questions for red balls and $3$ for green ones. 
I will take $\{a_1,a_2\}$, if the fake ball is in it, than using one more question I can know which one is it, if not than $a_3$ is the fake ball.
Here I will take $\{b_1,b_2,b_3\}$; if a fake ball is there, then using $2$ question I can  find it (like $\{a_1,a_2,a_3\}$). If not, then with one more  question I can find it in $\{b_4,b_5\}$.
 A: The possibility space for the fake balls is the cartesian product between the set $R$ of red balls and the set $G$ of green balls (you can visualise it as a $3 \times 5$ grid).
For a given question, let $A$ be the set of red balls that were not chosen, and $B$ the set of green balls that were not chosen. If the answer is "no", we know that the answer is in $A \times B$. If the answer is "yes", it is in its complement.
Each question divides the possibility space at most in half, so that, after the first question, if one of the sets has more than 8  (=$2^3$) elements, we lose.
Thus, for the first question, either $|A \times B| = 7$ or $|A \times B| = 8$. Since $|A \times B| = |A||B|$, it is easy to see that the only possibility is if $|A| = 2$ and $|B| = 4$, that is, if the first question involves exactly one red ball ($a_1$) and one green ball ($b_1$).
$$
\begin{array}{|c|c|c|c|c|}
\hline
Y & Y & Y & Y & Y \\ \hline
Y & n & n & n & n \\ \hline
Y & n & n & n & n \\ \hline
\end{array}
$$
If the answer is "no", we use one question to discover which of the two balls in $A$ is fake and the remaining two to discover which of the four balls in $B$ is fake.
If the answer is "yes", we use one question to discover if the green ball $b_1$ is fake. If yes, then we have two questions to resolve the three red balls. If no, then the first red ball $a_1$ is fake, and we have two questions to resolve the other four green balls in $B$.
$$
\begin{array}{|c|c|c|c|c|}
\hline
Y & n & n & n & n \\ \hline
Y & & & & \\ \hline
Y & & & & \\ \hline
\end{array}
$$
