$X \sim\operatorname{Exp}(1/2), Y \sim\operatorname{ Exp}(1) $ Given $Z$ find $E[Y|Z], V[Y|Z]$ $X \sim \operatorname{Exp}(1/2), Y \sim \operatorname{Exp}(1) $ they are independent. find $E[Y|Z], V[Y|Z]$ where $Z = -1$ if $X < Y$ and $0$ otherwise.
$\displaystyle P(Y\le y \mid Z = 1) = \frac{P(Y \le y, X < Y) }{P(X < Y)}$.
$\displaystyle P(X < Y) = \frac{1/2}{1/2+1} = \frac{1}{3}$
The other case would be similar.
The problem is trying to find $P(Y \le y, X < Y) $ and the other case. 
$\displaystyle E[Y\mid Z] = \int_0^\infty y \bigg (\frac{P(Y \le y, X<Y)}{1/3} + \frac{P(Y \le y, X\ge Y)}{2/3}\bigg ) dy$
 A: As you also found, 
\begin{align}
P(X<Y)&=\iint\mathbf1_{x<y}f_X(x)f_Y(y)\,\mathrm{d}x\,\mathrm{d}y
\\&=\frac12\iint\mathbf1_{0<x<y}\,e^{-x/2}e^{-y}\,\mathrm{d}x\,\mathrm{d}y
\\&=\frac12\int_0^\infty e^{-y}\left(\int_0^y e^{-x/2}\,\mathrm{d}x \right) \mathrm{d}y
\\&=\frac13
\end{align}
For the conditional expectation, proceed as
\begin{align}
E\,[Y\mid Z=-1]&=E\,[Y\mid X<Y]
\\&=\frac{E\,[Y\mathbf1_{X<Y}]}{P(X<Y)}
\\&=3\iint y\mathbf1_{x<y}f_X(x)f_Y(y)\,\mathrm{d}x\,\mathrm{d}y
\\&=\frac32\iint y\mathbf1_{0<x<y}\,e^{-x/2}e^{-y}\,\mathrm{d}x\,\mathrm{d}y
\\&=\frac32\int_0^\infty y e^{-y}\left(\int_0^y e^{-x/2}\,\mathrm{d}x\right)\mathrm{d}y
\\&=\frac53
\end{align}
You can find $E\,[Y\mid Z=0]$ similarly.
Or if you want you can use the law of total expectation:
$$E\,[Y]=E\,[Y\mid Z=-1]P(Z=-1)+E\,[Y\mid Z=0]P(Z=0)$$
So you have $$1=\frac53\cdot\frac13+E\,[Y\mid Z=0]\cdot\frac23$$
That is, $$E\,[Y\mid Z=0]=\frac23$$
Combining the two expectations we have $$E\,[Y\mid Z=z]=\frac23-z\quad,\,z\in\{0,-1\}$$
So your answer should say $$\color{green}{E\,[Y\mid Z]=\frac23-Z}$$
I would do the same for $E\,[Y^2\mid Z]$ and hence find $\operatorname{Var}[Y\mid Z]=E\,[Y^2\mid Z]-(E\,[Y\mid Z])^2$.
