Two fair dice are rolled. Find the joint probability mass function of X and Y...

Define ${\displaystyle N_1}$ and ${\displaystyle N_2}$ as random variables that mark numbers obtained on the first and the second die. We know that ${\displaystyle N_1}$ and ${\displaystyle N_2}$ are independent and that ${\displaystyle N_1,N_2}$ 
a) Here we have that ${\displaystyle X=max(N_1,N_2)}$ and ${\displaystyle Y=N_1+N_2}$. Thus ${\displaystyle X\in \{1,\dots ,6\}}$ and ${\displaystyle Y\in \{2,\dots ,12\}}$. Also we have that X<Y almost certainly. So take any ${\displaystyle k<l}$, where k and l are from the ranges given above. Consider event ${\displaystyle X=k,Y=l}$. That menas that the maximum value on any die is k and that the sum on both dice is l. Observe that the only possible pairs of ${\displaystyle (N_1,N_2)}$ corresponding to that event are ${\displaystyle (k,l,-k)}$ and ${\displaystyle (l,-k,k)}$ if ${\displaystyle l<2k}$. If ${\displaystyle l=2k}$, the only possible pair s (k,k). Hence the required PMF is 
$P(X=k,Y=l)=\{\begin{array}{ll}\frac{2}{36}& ,k<l<2k\\ \frac{1}{36}& ,l=2k\end{array}$

Here we have that $X=min(N_1,N_2)$ and $Y=max(N_1,N_2)$. We also have that $X\le Y$ almost certainy. So, take any $k\le l$
Suppose that k<l. In this case we have to have $N_1=k,N_2=l$ or $N_1=l,N_2=k$. So, there are only two possibilities, hence
$P(X=k,Y=l)=\frac{2}{36}$
if k=l, the only possibility is $(N_1,N_2)=(k,k)$, thus
$P(X=k,Y=l)=\frac{1}{36}$