Let $X,Y,Z$ to be discrete random variables with the property that their values are distinct with probability $1$. Let $a = P(X > Y), \ b = P(Y > Z), c = P (Z>X)$.
(a) Show that $\min\{a,b,c \} \leq 2/3.$
I do not know if this is the correct reasoning, but this is what i think of the problem so far. The sum of $a,b,c$ is $$ P(X > Y) + \ P(Y > Z) + P (Z>X) = E(I_{X > Y} + I_{Y >Z} + I_{Z > X}).$$ I think this sum is two because if such $t$ happens to be in $I_{X > Y}, \ I_{Y >Z}$, then it cannot be in $I_{Z < X}$. Then two out of three are satisfied and $\min\{a,b,c \} \leq 2/3$ is achieved.
(b) If $X,Y,Z$ are independent and identically distributed, then $a = b = c = 1/2$.
Since there are identically distributed, we have for example $P(X > Y) = P(Y > X)$. Also, $P(Z > X) = P(X > Z)$ and $P(Y >Z) = P(Z > Y).$ Does this means that we can use independence to get $$P(X > Y) + P(Z > X) = 1,$$ so $P(X > Y) = 1/2 = P(Z > X)$, but $P(X > Y) = P(Y > Z)$, so $a =b=c=1/2$.
I can feel that i am doing something wrong, but maybe is that i am not understanding the problem at all.
The use of $E(I_{X > Y} + I_{Y >Z} + I_{Z > X})$ is a good idea. As you observed, this value cannot be greater than $2$ (because $I_{X > Y} + I_{Y >Z} + I_{Z > X} \leq 2$).
If $X,Y,Z$ are independent and identically distributed discrete variables, then $a < \frac12$. You can make sense of the question if you drop the word "discrete", allowing continuous variables to be used. Then you have $$P(X>Y)=P(Y>X)$$ as you observed, but also $$P(X=Y) = 0.$$ Now apply total probability to the events $X>Y$, $X<Y$, and $X=Y$ and you should easily be able to show that $a = P(X>Y)=\frac12$.
Notice that you don't need to use variable $Z$ at all to get the value of $a$, but you can repeat the argument twice more (or simply invoke the symmetry of the problem statement) to solve for $b$ and $c$.