(a): 0; (b): No

The lesson: This problem is supposed to drive in the fact that zero correlation does not mean that there's no relationship between 2 variables. It's true that if you have no relationship, you will have zero correlation, but that rule only goes one way.

Part a

\[\begin{align}\text{Thrm 2.2.2}\\
Cov[X,Y] &= E[XY] - E[X]E[Y] \\
&= E[X(X^2)] - E[X]E[X^2]\\
\\
\text{Thrm 2.1.1}\\
E[X] &= \sum_x xf(x)\\
&= (-2)0.2 + (-1)0.2 + (0)0.2 + (1)0.2 + (2)0.2 \\
&= 0\\
\\
\text{Thrm 2.1.5} & ; \text{(LOTUS)} \\
E[X^2] &= \sum_x g(x)f(x)\\
&= (4)0.2 + (1)0.2 + (0)0.2 + (1)0.2 + (4)0.2 \\
&= 2\\
\\
E[X^3] &= \sum_x g(x)f(x)\\
&= (-8)0.2 + (-1)0.2 + (0)0.2 + (1)0.2 + (8)0.2 \\
&= 0\\
\\
\text{Plug in}\\
Cov[X,Y] &= 0 - (0)2\\
&= 0\end{align}\]

Part b)

We can just note that by inspection that $Y$ is dependent on $X$. Specifically, $Y$ is equal to $X^2$, so not only do we know that they are not independent, we know the exact function that makes $Y$ dependent on $X$.

Alternatively, we can show lack of independence using Definition 1.3.16 from the book. That definition says that variables $X$ and $Y$ are independent if their joint probability distribuion $f(x,y)$ and the individual marginal distributions $f_Y(y)$ and $f_X(x)$ are related by the equation: $f(x,y) = f(x)f(y)$. We can see below that this is not how they are related in our joint distribution function.

Table of joint distribution function $f(x,y)$