Calculating the ph of an unknown solution by measuring the precipitate
You have to assume your unknown hydroxide is completely dissociate in aqueous solution. If so, then you can calculate $\mathrm{pOH}$ of the solution as $\mathrm{pOH} = -\log \ce{[OH-]}$. Then, $\mathrm{pH} = \mathrm{p}K_\mathrm{w} - \mathrm{pOH} = \mathrm{p}K_\mathrm{w} +\log \ce{[OH-]}$. Now, you see, since $\mathrm{p}K_\mathrm{w} = 14$, only unknown is $\ce{[OH-]}$. 

Now look at the reaction of your unknown hydroxide with lead (II) nitrate:
$$\ce{2 OH- (aq) + Pb(NO3)2 (aq) -&gt; Pb(OH)2 (s) + 2 NO3 (aq)}$$

Since you know the precipitated amount of $\ce{Pb(OH)2}$ isolated from $\pu{1.0 L}$ of unknown hydroxide solution, you can calculate original $\ce{[OH-]}$:

$$\text{Amount of } \ce{OH-}\text{ in original solution} \\
= \pu {3.81 g}\text{ of } \ce{Pb(OH)2} \times \frac{\pu {1 mol}\text{ of } \ce{Pb(OH)2}}{\pu {241.21 g}\text{ of } \ce{Pb(OH)2}} \times \frac{\pu {2 mol}\text{ of } \ce{OH-}}{\pu {1 mol}\text{ of } \ce{Pb(OH)2}} \\
= \pu {0.0316 mol}\text{ of } \ce{OH-} $$ 
Thus, $ \ce{[OH-]} = \pu{0.0316 M}$, and hence,
$$\mathrm{pH} = \mathrm{p}K_\mathrm{w} +\log \ce{[OH-]}= 14 + \log(0.0316) = 12.5 $$

Asking is approximate $\mathrm{pH}$ of orignal solution. Thus, this answer is acceptable.  However, keep in mind that trace of $\ce{OH-}$ will remain in solution due to the solubility of  $\ce{Pb(OH)2}$ in water $\left(K_\mathrm{sp} = 1.42 \times 10^{-20}\right)$.