Cards game and probabilities 
From a standard 52-cards deck we randomly pull out 2 cards.
  One is black and the other is a 4.
  We choose one of the two cards.
  What is the probability:
  
  
*
  
*The card we pick is a 4
  
*The card we pick is black
  
*The card we pick is a face card

For 1: I believe it must be 1/2 (since 1 of the 2 cards is a 4) plus the probability for the other card also to be a 4. Since the other card is black, we know there are 2 fours in 26 black cards so 2/26? So finally $1/2+2/26$? Not sure.


*Similarly, 1/2 + the probability for the "4" card to be black - it can't be $1/2+2/4$ though.

*The black card can be a face card with probability $6/26$.
Can you help me out?
 A: In this answer I preassume that "one is black" stands for "at least one is black".
1)
This question can be rephrased as: "if we take out randomly one by one two cards without replacement then - if one of the cards taken out appears to be black and the other a $4$ - what is the probability that the first card taken out is a $4$.
Let $E$ denote the event that the first drawn card is a $4$.
Let $A$ denote the event that two $4$'s are drawn of different color.
Let $B$ denote the event that two black $4$'s are drawn.
Let $C$ denote the event that a  black card that is not a $4$ is drawn and a $4$.
Then:$$P\left(E\mid A\cup B\cup C\right)=\frac{P\left(E\cap A\right)+P\left(E\cap B\right)+P\left(E\cap C\right)}{P(A)+P\left(B\right)+P\left(C\right)}=$$$$=\frac{4\cdot2+2\cdot1+4\cdot25}{4\cdot2+2\cdot1+2\cdot4\cdot25}=\frac{11}{21}$$

Edit:
The comment of Oldboy (thank you) on this stimulated me to have a second look and I found a mistake in my answer (my apologies for that) The $25$ must be (of course) $24$ (the number of black cards that are not a $4$). This leads to the outcome:$$\frac{4\cdot2+2\cdot1+4\cdot24}{4\cdot2+2\cdot1+2\cdot4\cdot24}=\frac{106}{202}=\frac{53}{101}\approx0.524752475$$
This agrees with the outcome of Oldboy.
To make things a bit more clear:
$$P\left(\text{one black and other }4\right)=$$$$P\left(2\text{ black }4\text{'s}\right)+P\left(\text{black }4\text{ and red }4\right)+P\left(\text{one }4\text{ and one black that is not a }4\right)=$$$$\frac{2}{52}\frac{1}{51}+2\frac{2}{52}\frac{2}{51}+2\frac{4}{52}\frac{24}{51}$$
is the denominator, and:$$P\left(\text{one black and other }4\text{ and first one is a }4\right)=$$$$P\left(2\text{ black }4\text{'s}\right)+P\left(\text{black }4\text{ and red }4\right)+P\left(\text{first a }4\text{ and then a black that is not a }4\right)=$$$$\frac{2}{52}\frac{1}{51}+2\frac{2}{52}\frac{2}{51}+\frac{4}{52}\frac{24}{51}$$
is the numinator.

Edit2 (I decided to provide the other answers as well)
2) 
Here the denominator is the same as in 1) and the numerator is:$$P\left(\text{one black and other }4\text{ and first one is black}\right)=$$$$P\left(\text{first a black }4\text{ then a black card or a red }4\right)+P\left(\text{first a black that is not a }4\text{ and then a }4\right)=$$$$\frac{2}{52}\frac{27}{51}+\frac{24}{52}\frac{4}{51}$$leading to probability $$\frac{150}{202}=\frac{75}{101}$$

3)
I preassume that $J,Q,K$ are face cards and the aces are not.
Here the denominator is the same as in 1) and the numerator is:$$P\left(\text{one black and other }4\text{ and first one is face}\right)=P\left(\text{first is a black face and second is a }4\right)=\frac{6}{52}\frac{4}{51}$$leading to probability:$$\frac{24}{202}=\frac{12}{101}$$
