The probability mass function $ P $ of a discrete random variable $ X $ is given by $$ P(x) = \begin{cases} \dfrac{x^{2}}{a} & \text{if $ x \in \{ -2,-1,0,1,2 \} $}; \\ 0 & \text{otherwise}. \end{cases} $$
(a) Find $ a $ and $ \mathbf{E}[X] $.
I have no clue how to find $ a $. I believe that $ \mathbf{E}[X] $ is just the mean of $ \{ -2,-1,0,1,2 \} $, so would it be $$ \frac{(-2) + (-1) + 0 + 1 + 2}{5} = 0? $$
(b) What is the probability distribution of the random variable $ Z = (X - \mathbf{E}[X]) $?
I have no clue how to do this.
Find $a$ through the fact that $1 = \sum\limits_{x=-2}^2 P(x)$ by Law of Total Probability.
$$1 = \frac 1 a\sum\limits_{x=-2}^2 {x^2}$$
Then use, $\mathsf E(X) = \sum\limits_{x=-2}^2 x~P(x)$, which is the weighted average.
$$\mathsf E(X) = \frac 1 a\sum\limits_{x=-2}^2 x^3$$
Finally $\Pr(Z=z) $ is found by a simple Linear Transform.
$$\begin{align}\Pr(Z=z)~=~& \Pr(X-\mathsf E(X) = z) \\ =~& P(z+\mathsf E(X)) \\ \ddots& \end{align}$$
PS: You will also need to identify the support for $Z$.