How much calcium hydroxide will precipitate after addition of sodium hydroxide into saturated calcium hydroxide solution?
A less analytic aproach:

Initial concentrations: 

$K_{\mathrm{sp}}=\ce{[Ca^{2+}][OH^-]^2}=x\cdot (2x)^2=8\cdot10^{-6}$

$\ce{[Ca^{2+}]=0.0126}$ M; $\ce{[OH^-]=0.0252}$ M.

Now 0.1 mol NaOH is added. Let $p$ be the fraction that precipitates.
The concentrations that remain in solution are:

$\ce{[Ca^{2+}}]=0.0126(1-p)$

$\ce{[OH^-]}=(0.10 + 0.0252 -2\cdot0.0126p)^2$

Since the solubility product remains the same:

$K_{\mathrm{sp}}=0.0126(1-p)(0.1252 -2\cdot0.0126p)^2$

Now, since there are only four scenarios, try them out substituting every value of $p$ in  the equation and take the closest one to $8\cdot10^{-6}$:

\begin{array}{|c|c|}\hline
    p&amp;K_\mathrm{sp}\\\hline
    0.50&amp;\pu{7.99E-05}\\
    0.75&amp;\pu{3.56E-05}\\
    0.95&amp;\pu{6.46E-06}\\
    0.99&amp;\pu{1.27E-06}\\\hline
\end{array}

The closest one is for roughly 95% precipitation, without solving any equation.