For two random variables $X_1 + X_2$ and $\min(X_1,X_2)$ find the joint-distribution and the covariance Let $X_1,X_2$ be independent random variables.
Moreover $X_1,X_2$ are discrete uniform distributed({$1,...,N$})
We define:
$A:=  X_1+X_2$
$B:= \min(X_1,X_2)$
Find joint-distribution of $A$ and $B$  and calculate covariance of $A$ and $B$.
I don't really have a clue how to find the joint-distribution of $A$ and $B$. So for this task I really need some help. Maybe you can give me a hint and I try to solve it then. I would edit this question with my attempt until I find the joint-distribution.
Edit: Let us start with the joint distribution. $P(A=a,B=b) =P(B=b|A=a)\cdot P(A=a)$.
In the answers below I saw that: $ P(A=a)=\frac{1}{N^2}\left\{ \begin{array}{lr}a-1& 2\leq a \leq N+1 \\
2N+1-a & N+2\leq a \leq 2N \end{array} \right. $ 
I understand why this formula holds for $P(A=a)$ but only with the example. I don't know how we can show it. Moreover We have to find $P(B=b|A=a)$. I think a) is clear now. Will try to edit my attempt in a few days.
Edit for the second part: I will write $Cov$ for covariance.
So we have to calculate $Cov(A,B)$. 
We already know that: $Cov(A,B) = E(AB) - E(A)E(B) $ (expected value)
All we have to compute is the expected value for $AB,A,B$. Thanks to the user "mathemagical". I already know the value of $E(A), E(B)$. I even understand the rest of the answer except how we can calculate $E(Z)$.
 A: I am assuming that you mean the discrete uniform distribution that gives probability      $\frac{1}{N}$ to each point.  As the joint distribution is a bit gnarly, it seems easier to compute the covariance by computing the various pieces of $E(AB)-E(A)E(B)$. 
Hints/Outline: 
(For easy typing, I have changed $X_1$ to $X$ and $X_2$ to $Y$). 
For the covariance of $A$ and $B$
First find the distribution and expectation for A and B. Then do the same for $AB$.
From independence, the joint distribution of $(X,Y)$ is easy, with the probability associated with any $(x,y)$ pair given by $\frac{1}{N^2}$
Compute $E(A)$ and $E(B)$
Note that $A=X+Y$ takes values from 2 to 2N with probabilities as follows
$$P(A=k)=\frac{1}{N^2}\left\{ \begin{array}{lr}k-1& 2\leq k \leq N+1 \\
2N+1-k & N+2\leq k \leq 2N \end{array} \right.$$
(do you see why? The values taken by A for each X and Y when N=6 are shown in the table below, where X and Y are shown in the margins) 
$$\begin{array}{c|cccccc} 6&7&8&9&10&11&12\\
5&6&7&8&9&10&11\\
4& 5&6&7&8&9&10\\
3& 4& 5&6&7&8&9\\
2&3& 4& 5&6&7&8\\
1&2&3&4& 5&6&7\\
\hline
A&1&2&3&4& 5&6 \end{array}$$
This gives (derive )  $$E(A)=N+1$$
Note that $B=X$ when $X \le Y$ and $B=Y$ when $X\ge Y$ as in the example table below for $N=6$. 
$$\begin{array}{c|cccccc} 6&1&2&3&4&5&6\\
5&1&2&3&4&5&5\\
4&1&2&3&4&4&4\\
3&1&2&3&3&3&3\\
2&1&2&2&2&2&2\\\
1&1&1&1&1&1&1\\
\hline
B&1&2&3&4& 5&6 \end{array}$$
So (derive)
$$P(B=k)=\frac{2N-2k+1}{N^2}$$ and $$E(B)=\frac{N(N+1)(2N+1)}{6N^2}$$
Compute $E(AB)$
Note that
 $$AB=\left\{ \begin{array}{lr} XY + Y^2& Y\le X\\XY+X^2&Y>X\end{array}\right.$$ 
$$AB=XY +\min(X^2,Y^2)=XY+Z$$ 
Now note that $E(XY)=E(X)E(Y)=\left(\frac{N+1}{2}\right)^2$ by independence. You can now finish the computation of $Cov(A,B)=E(XY)+E(Z)-E(A)E(B)$ by finding $E(Z)$ (you can recycle the distribution of $B$ for this: note that $P(Z=k^2)=P(B=k)$)
That then leads you to the covariance of A and B since you already have $E(A), E(B)$.
For the joint cumulative distribution of  $A$ and $B$, use the geometry of the tables for $A$ and $B$ to note that there are 3 cases (and some meticulous computation for the probabilities in each of the 3 cases). 
$$F(a,b)\equiv P(A\le a, B\le b)=\frac{1}{N^2} \left \{\begin{array}{lr}
2Nb-b^2& a\ge N+b\\
N^2-\frac{(2N-a)(2N-a+1)}{2}&a\le 2b\\
N^2-\frac{(2N-a)(2N-a+1)}{2} - \frac{(a-2b)(a-2b+1)}{2}& \mbox{otherwise} \end{array}\right.$$
If you want to find the probability mass function, then $$P(A=a,B=b)=F(a,b)-F(a-1,b)-F(a,b-1)+F(a,b)$$
A: Hint
I think the fastest progress will be to calculate the joint distribution directly (since you're asked for that anyway) and then use that to find the covariance via the standard formula. 
You can get the joint using combinatorics. The simplest thing to directly compute is probably $$ P(B=b\mid A = a).$$ For instance conditional on $A=4,$ it is equally likely for $(X,Y)$ to be $(1,3)$ $(2,2)$ or $(3,1)$ so $B$ conditionally has a $1/3$ probability of being $2$ and a $2/3$ probability of being $1.$
