Total Conditional Probability and two urns with different number of balls 
Let E, F, G strictly positive probability events.

*

*(A.) Prove the following formula of the total probability conditioning
$$P(E|G) = P(E|F ∩ G)P(F|G) + P(E|F^c ∩ G)P(F^c|G).$$

*On the table, there are two urns, one containing $1$ red ball and $n-1$ black and the other containing $1$ black ball and $n-1$ red. You chose a random urn and extracted a ball, therefore, without replacement, it is extracted a second ball from the same urn. Calculate the conditional probability of "the second ball extracted is black" gave the event "the first is red".


The first thing I have done is to decompose the formula in
$$P(E|G) = \frac{P(E ∩ F ∩ G)}{P(F ∩ G)}\frac{P(F ∩ G)}{P(G)} + \frac{P(E ∩ F^c ∩ G)}{P(F^c ∩ G)}\frac{P(F^c ∩ G)}{P(G)}$$
and then
$$P(E|G) = \frac{P(E ∩ F ∩ G)}{P(G)} + \frac{P(E ∩ F^c ∩ G)}{P(G)}$$
$$P(E|G) = \frac{1}{P(G)}[P(E ∩ F ∩ G)P(G) + P(E ∩ F^c ∩ G)]$$
but I don't know how to go forward.
About the second point, I thought I have to use this formula to resolve it but I can't find how. I have only understood that because we have the two urns, we have to do $0,5 * ...$ to find the probability
 A: Just a presentation note: you shouldn't write your working as
$$P(E|G) = \frac{P(E ∩ F ∩ G)}{P(F ∩ G)}\frac{P(F ∩ G)}{P(G)} + \frac{P(E ∩ F^c ∩ G)}{P(F^c ∩ G)}\frac{P(F^c ∩ G)}{P(G)}$$
until you have actually proved this is true. Work with one side and find a way to turn it into the other, rather than attempt to end with a line such as
$$P(E|G)=P(E|G).$$
On to the actual proof, I'll continue from your second line. We have
$$P(E|F\cap G)P(F|G)+P(E|F^c\cap G)P(F^c|G)=\frac1{P(G)}(P(E\cap F\cap G)+P(E\cap F^c\cap G))\\
=\frac1{P(G)}P(E\cap G)=P(E|G).$$
For the second question, you are correct to use this formula. The events in question will be $E=$ "the second ball is black", $G=$ "the first ball is red", and $F=$ "the urn you are drawing from contains $1$ red ball". The hardest part is calculating $P(F|G)$. We use
$$P(F|G)=\frac{P(F\cap G)}{P(G)}=\frac{P(F)P(G|F)}{P(G)}$$
We have $P(F)=\frac12$ (each urn is equally likely to be drawn from), $P(G)=\frac12$ (there are a total of $n$ balls of each color), and $P(G|F)=\frac1n$ (you are trying to choose the $1$ red ball out of $n$ total balls), so $P(F|G)=\frac1n$. This implies $P(F^c|G)=\frac{n-1}n$.
Finally, $P(E|F\cap G)=1$ (since there are no red balls left) and $P(E|F^c\cap G)=\frac{n-2}{n-1}$ (since there are $n-1$ balls left and all bar one are red). Putting this all together we obtain
$$P(E|G)=1\cdot\tfrac1n+\tfrac{n-2}{n-1}\cdot\tfrac{n-1}n=\tfrac{n-1}n.$$
