Tower Property With Product of 2 Distributions Problem: $Z\sim \operatorname{Ber}(.55)$ and $X\sim \exp(\frac{1}{8}) $ or exp with mean 8 and variance 64.
Calculate $\operatorname{Var}[XZ]$
The solution offers the guidance of using the property $Y=XZ$ for:
$\operatorname{Var}[Y]=E[\operatorname{Var}[Y\mid Z]]+\operatorname{Var}[E[Y\mid Z]]$.
When $Z=0$, $E_Y[Y\mid Z]=\operatorname{Var}_Y[Y\mid Z]=0$ and when $Z=1$, $Y=X  \implies E_Y[Y\mid Z=1]=E_X[X]=8$ and  $\operatorname{Var}_Y[Y\mid Z=1]=\operatorname{Var}_X[X]=64$. This part makes perfect sense to me. Here is the part that doesn't make sense to me. The solution goes on to say that $E_Y[Y\mid Z]=E_X[X]=8Z$ and $\operatorname{Var}_Y[Y\mid Z]=\operatorname{Var}_X[X]=64Z$, so my main question is, why is there a coefficient  $Z$ now  when previously we just had a number??
 A: 
The solution goes on to say that $E_Y[Y\mid Z]=E_X[X]=8Z$

That is not quite correct (the middle expression is erroneous).
  Assuming that $Z$ is independent from $X$, then so too will $X$ be conditionally independent from $Z$ when given $Z$, and therefore:-
$$\begin{align}\mathsf E(Y\mid Z)&=\mathsf E(XZ\mid Z)\\&=\mathsf E(X\mid Z)\cdot\mathsf E(Z\mid Z)\\&=\mathsf E(X)\cdot Z\\&=8\,Z\end{align}$$
Likewise for the variance:
$$\begin{align}\mathsf{Var}(Y\mid Z)&=\mathsf {Var}(XZ\mid Z)\\&=\mathsf E(X^2Z^2\mid Z)-\mathsf E(XZ\mid Z)^2\\&=\mathsf E(X^2\mid Z)\cdot\mathsf E(Z^2\mid Z)-\mathsf E(X\mid Z)^2\cdot\mathsf E(Z\mid Z)^2\\&=\mathsf E(X^2)\cdot Z^2-\mathsf E(X)^2\cdot Z^2\\&=\mathsf{Var}(X)\cdot Z^2\\&=64\,Z^2\end{align}$$


why is there a coefficient $Z$ now when previously we just had a number?

Previously you were conditioning on the event of $Z$ being a specified value.  Here you are conditioning over the random variable $Z$ being an unspecified value (it may be whatever it may be). These are related concepts.
So $\mathsf {Var}(Y\mid Z{=}0)~{=\left.64\, Z^2\right\rvert_{\small Z=0}\\=0}\\\mathsf {Var}(Y\mid Z{=}1)~{=\left.64\,Z^2\right\rvert_{\small Z=1}\\=64}$

Thus $\mathsf E(\mathsf{Var}(Y\mid Z))~{=64\,\mathsf E(Z^2)\\=64\cdot(0^2\cdot 0.45+1^2\cdot 0.55)\\=35.2}$
And $\mathsf{Var}(\mathsf{E}(Y\mid Z))~{=\mathsf{Var}(8\,Z)\\=64\cdot(0.55(1-0.55))\\=15.84}$
A: It might be more clear to write it out as: $E_Y[Y\mid Z = 0]=\operatorname{Var}_Y[Y\mid Z = 0]=0$ and $E_Y[Y\mid Z = 1]= E_X[X] = 8 $ and finally:
$$\operatorname{Var}_Y[Y\mid Z = 1]=\operatorname{Var}_X[X] = 64$$
What we really want is $\operatorname{Var}[Y] = \operatorname{Var}[Y]=E[\operatorname{Var}[Y\mid Z]]+\operatorname{Var}[E[Y\mid Z]]$. This is just $\mathbb{E}[64Z] + \operatorname{Var}[8Z]$ which can be evaluated with what you have above.
$Z$ appears in the conditional expectation since when you condition on $Z$, you treat it as a constant.
