= Solution
For $0<x<1$, the conditional density is obtained by dividing the joint density by the positive marginal density:
$$
f_{Y\mid X=x}(y)=\frac{24xy}{12x(1-x)^2}
=\frac{2y}{(1-x)^2},\qquad0\leq y\leq1-x.
$$
Consequently
$$
\boxed{f_{Y\mid X=1/2}(y)=\begin{cases}8y,&0\leq y\leq1/2,\\0,&\text{otherwise},\end{cases}}
$$
which integrates to one. Its <conditional expectation> is
$$
\boxed{\mathbb E(Y\mid X=1/2)=\int_0^{1/2}y(8y)\,dy=\frac13.}
$$
This uses the usual continuous conditional-density version, defined by the density ratio at a point where $f_X$ is positive; conditioning does not require the event $X=1/2$ itself to have positive <probability>.
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