Let $X$ be a random variable with pdf $f(x)$. Let $Y$ be defined as follows: $$ Y = \begin{cases} \begin{align} &0,\qquad &X \le A\\ &X - A,\qquad &A \lt X \le L\\ &L - A,\qquad &X \gt L \end{align} \end{cases} $$
$E(Y|Y>0)$ should be: $$ \frac{\int_A^L (x-A) f(x)\;dx + (L-A) \int_L^\infty f(x)\;dx}{\int_A^\infty f(x)\;dx} $$
Am I correct in my assumption that the variance of $Y$ conditional on $Y > 0$ is $E(Y^2|Y > 0) - (E(Y|Y > 0) ) ^ 2$ where $E(Y^2|Y > 0)$ is:
$$
\frac{\int_A^L (x-A)^2 f(x)\;dx + (L-A)^2 \int_L^\infty f(x)\;dx}{\int_A^\infty f(x)\;dx}
$$
in other words, is Var$(Y|Y > 0)$ equal to:
$$
\frac{\int_A^L (x-A)^2 f(x)\;dx + (L-A)^2 \int_L^\infty f(x)\;dx}{\int_A^\infty f(x)\;dx} - \left(\frac{\int_A^L (x-A) f(x)\;dx + (L-A) \int_L^\infty f(x)\;dx}{\int_A^\infty f(x)\;dx}\right)^2
$$
Similarly, for the case: $$ Z = \begin{cases} \begin{align} &0,\qquad &X \le A\\ &X - A,\qquad &A \lt X \le L\\ &0,\qquad &X \gt L \end{align} \end{cases} $$
$E(Z|Z > 0)$ would be $$ \frac{\int_A^L (x-A) f(x)\;dx}{\int_A^L f(x)\;dx} $$
and would Var$(Z|Z > 0)$ then equal: $$ \frac{\int_A^L (x-A)^2 f(x)\;dx}{\int_A^L f(x)\;dx} - \left(\frac{\int_A^L (x-A) f(x)\;dx}{\int_A^L f(x)\;dx}\right)^2 $$
Thank you.
Your expressions for $\mathbb{E}[Y|Y>0]$ and $\operatorname{var}[Y|Y>0]$ are correct. They are immediate consequences of the fact that $Y>0$ implies $X>0$ and standard results on the distribution of a truncated random variable. See, e.g. the wikipedia page on truncated distributions.
Likewise, your expressions for $\mathbb{E}[Z|Z>0]$ and $\operatorname{var}[Z|Z>0]$ are both correct too, as $Z>0$ implies $A<X<L$.