Our mini-deck has $12$ cards. There are $\binom{12}{4}$ equally likely ways to choose $4$ cards from this deck.
Now we count the favourables, the hands that have $2$ Queens and $2$ Kings.
The $2$ Queens can be chosen from the $4$ available in $\binom{4}{2}$ ways. For each way of choosing $2$ Queens, there are $\binom{4}{2}$ ways to choose the $2$ Kings. Thus the total number of favourables is $\binom{4}{2}\binom{4}{2}$.
For the probability, divide.
For the second problem, there are $\binom{4}{2}$ ways to choose the $2$ Queens. Once this is done, the Kings are determined, since they must be of the same suits as the chosen Queens. Thus the number of favourables in the "suits must match" problem is just $\binom{4}{2}$.
Split up like $52=4+4+44$, i.e. $4$ kings $4$ aces and the other $44$ cards.
In total $N:=\binom{52}2$ different pair-picks are possible and equiprobable.
- $\binom40\binom40\binom{44}2$ of them give $(K,A)=(0,0)$
- $\binom40\binom41\binom{44}1$ of them give $(K,A)=(0,1)$
- $\binom41\binom40\binom{44}1$ of them give $(K,A)=(1,0)$
- $\binom40\binom42\binom{44}0$ of them give $(K,A)=(0,2)$
- $\binom41\binom41\binom{44}0$ of them give $(K,A)=(1,1)$
- $\binom42\binom40\binom{44}0$ of them give $(K,A)=(2,0)$
You can find the probabilities by dividing the outcomes by $N$.
edit (alternative):
Pick two cards one by one.
For $i=1,2$ let $A_{i},K_{i},E_{i}$ denote the events that the $i$-pick
gives ace, king and other card respectively.
$P\left(A_{1}\cap A_{2}\right)=P\left(A_{1}\right)P\left(A_{2}\mid A_{1}\right)=\frac{4}{52}\frac{3}{51}$
$P\left(A_{1}\cap K_{2}\right)=P\left(A_{1}\right)P\left(K_{2}\mid A_{1}\right)=\frac{4}{52}\frac{4}{51}$
$P\left(A_{1}\cap E_{2}\right)=P\left(A_{1}\right)P\left(E_{2}\mid A_{1}\right)=\frac{4}{52}\frac{44}{51}$
$P\left(K_{1}\cap A_{2}\right)=P\left(K_{1}\right)P\left(A_{2}\mid K_{1}\right)=\frac{4}{52}\frac{4}{51}$
$P\left(K_{1}\cap K_{2}\right)=P\left(K_{1}\right)P\left(K_{2}\mid K_{1}\right)=\frac{4}{52}\frac{3}{51}$
$P\left(K_{1}\cap E_{2}\right)=P\left(K_{1}\right)P\left(E_{2}\mid K_{1}\right)=\frac{4}{52}\frac{44}{51}$
$P\left(E_{1}\cap A_{2}\right)=P\left(E_{1}\right)P\left(A_{2}\mid E_{1}\right)=\frac{44}{52}\frac{4}{51}$
$P\left(E_{1}\cap K_{2}\right)=P\left(E_{1}\right)P\left(K_{2}\mid E_{1}\right)=\frac{44}{52}\frac{4}{51}$
$P\left(E_{1}\cap E_{2}\right)=P\left(E_{1}\right)P\left(E_{2}\mid E_{1}\right)=\frac{44}{52}\frac{43}{51}$
As an example note that e.g. $$P(A=1,K=1)=P(A_1\cap K_2)+P(K_1\cap A_2)=\frac{4}{52}\frac{4}{51}+\frac{4}{52}\frac{4}{51}$$
Best Answer
For $(i)$ and $(ii)$ you may find it simpler to directly multiply probabilities.
$(i): \dfrac4{52}\cdot\dfrac3{51}\cdot\dfrac2{50}\cdot\dfrac1{49}$
$(ii): \dfrac{48}{52}\cdot\dfrac{47}{51}\cdot\dfrac{46}{50}\cdot\dfrac{45}{49}$
$(iii):$ Ok, by linearity of expectation, $4\times\dfrac4{52} = \dfrac4{13}$