I'm not very good at statistic and is given the problem this problem:

Here's the work that I've done:

I was wondering if I should consider all the dots, 21 dots, for the probability for part (iv) to get 9/21 for P(x|A) or just consider the dots where y<x to get my current answer of 9/9?
Also the same thing for part (v), should the probability be ___/21 or ___/9? Any help would be appreciated. Thanks in advance!
Not quite. It is $\mathsf P(X{\in}A)$ that equals $\sum_{x\in A}\mathsf P(X{=}x)$. The conditional probability is then:
$$\begin{align}\mathsf P(X=x\mid X>Y)&=\dfrac{\mathsf P(X=x, Y<x)}{\mathsf P(X>Y)}\\&=\tfrac 19\mathbf 1_{x=1}+\tfrac 29\mathbf 1_{x=2}+\tfrac 39\mathbf 1_{x=3}+\tfrac 29\mathbf 1_{x=4}+\tfrac 19\mathbf 1_{x=5}\end{align}$$
However that you had the correct probabilities in the calculation of your expected value.
Also note how your graphic indicates that the mass of the triangle centres on $X=3$.