Calculating the pH of a solution of Ca(OH)2
You did not account for the fact that $\ce{Ca(OH)2}$ is partly precipitated. Firstly, to find out the moles of $\ce{Ca(OH)2}$ precipitated:-
The number of moles in solution is $\frac{1.48}{74}M$ which is $0.02M$. The total moles precipitated is just the moles of $\ce{CaO}$ minus $0.02M$ since the initial number of moles of 
$\ce{Ca(OH)2}$ is equal to the moles of $\ce{CaO}$ taken and hence the amount precipitated turns out to be $0.03M$. Nevertheless, this is not required. With $0.02M$ $\ce{Ca(OH)2}$ in solution, the $\ce{OH-}$ concentration turns out to be $0.04M$ which, assuming no contribution of $\ce{OH-}$ from water and complete ionization, gives a $p\ce H$of $12.6$.