Expectation and variance of $Y = (X+2)^3$ where $X \sim \text{Ber}(p)$ I am confused about how to calculate the expected value and variance of this Bernoulli Random Variable $Y = (X+2)^3$ where $X \sim \text{Ber}(p)$. Currently I understand that the expected value of $X^2$ is the also $p$ but I am not understanding what I can do with that, or if I can do anything with it, to answer my question.

You seem confused about what $Y = (X+2)^3$ means.  Together with $X \sim Ber(p)$ they simply mean this:

*

*With probability $p$, we have $X = 1$, and therefore $Y = (X+2)^3 = 27$


*With probability $1-p$, we have $X = 0$, and therefore $Y = (X+2)^3 = 8$
Therefore, $Y$ is a bi-valued variable, taking value $27$ with prob $p$, and value $8$ with prob $1-p$.  From first principles, we have:
$$E[Y] = 27p + 8(1-p) = 19p + 8$$
$Var(Y)$ can also be calculated from first principles, i.e.
$$Var(Y) = E[(Y - E[Y])^2] = p(27 - (19p+8))^2 + (1-p)(8 - (19p+8))^2= 19^2 p (1-p)$$
The variance has such a nice form, because it is a rescaled Bernoulli variable.  As you may know, the standard Bernoulli ($\{0,1\}$-valued) has variance $p(1-p)$, and variance is shift invariant and $Var(cZ) = c^2 Var(Z)$ for any r.v. $Z$ and any constant $c$.  So in this example, the only thing that mattered is the gap between $8$ and $27$, which is $19$, and the variance is $19^2 \times$ that of a standard Bernoulli.
If you really want to use "formulas", then you can proceed as follows:
$$\begin{align}E[Y] &= E[(X+2)^3] \\&= E[X^3 + 6X^2 + 12X + 8] \\&= E[X^3] + 6E[X^2] + 12 E[X] + 8 \\&= 1p + 6p + 12p + 8 \\& = 19p + 8\end{align}$$
Here we used linearity of expectations, and also the facts $E[X^3] = E[X^2] = E[X] = p$.  However for this problem, this is overkill.  Also, deriving $Var(Y)$ this way would be:
$$Var(Y) = E[Y^2] - E[Y]^2 = E[(X+2)^6]- (19p + 8)^2$$
and you can expand the sixth power and use the facts $E[X^k] = p$ for any $k>1$ when $X$ is standard Bernoulli.
