Proof that Corr(X,Y) equals zero for uniform discrete R.V. i) X is a discrete uniform r.v. on the set $\{-1,0,1\}$. Let $Y=X^2$ . 
Prove that $Corr(X,Y)=0$.
ii) X is a discrete uniform r.v. on the set $\{-1,0,1\}$. Let $Y=X^2$. Are X and Y independent?
Confused on the fact that this uniform discrete RV has 3 parameters when Im use to seeing it with only 2. So for example to the the expected value it's just: $\frac{a+b}{2}$; however, here we have more than two options for "a" and "b". This is all under the assumptions that I can calculate E(X) and $E(Y)$; $Var(X)$ and $Var(Y)$; all to find $Cov(X,Y)$ and then $Corr(X,Y)$
 A: Uniform distribution over a finite set $\{x_1,\ldots,x_n\}$ just means $\Pr(X=x_i)=\frac{1}{n}$ for each $i$.
For your claims, with $Y=X^2$
$$
\text{Cov}(X,Y)=E(XY)-\underbrace{E(X)}_0E(Y)=E(X^3)=\frac{1}{3}1^3+\frac{1}{3}0^3+\frac{1}{3}(-1)^3=0.
$$
So $X$ and $Y$ are uncorrelated. Nonetheless, knowing $X$ gives $Y$ so $X$ and $Y$ are not independent. For example, you can note
$$
\Pr(X=1,Y=0)=\Pr(X=1,X^2=0)=0\neq\underbrace{\Pr(X=1)}_{1/3}\underbrace{\Pr(Y=0)}_{1/3}.
$$
A: 
Confused on the fact that this uniform discrete RV has 3 parameters when Im use to seeing it with only 2. 

What you are used to seeing is a uniform discrete distribution over an integer interval with upper and lower bounds, say $a$ and $b$.  Like so:
$$X\sim \mathcal{U}\{a..b\}$$
While some authors do use a comma rather than elipsis, what is always meant is a shorthand for the interval of all integers between (and including) the lower and upper bounds:
$$X\sim\mathcal{U}\{a, a+1, \ldots, b-1, b\} $$ 
Now, the set $\{-1, 0, 1\}$ does have integer steps between members, so could be represented as $\{-1..1\}$.   The questioneer may have felt that the set was small enough not to need the short hand, or just wanted to be extra clear that the set had those three members.
