Estimate the concentration of HCl and volume of 6 M HCl required to complete precipitate each of the given salts
Your calculations are looking good. I didn’t check the numerical values, I’ll just assume you’re able to do your algebra and use a calculator correctly.

What you got are two different concentrations required to start to precipitate all the cations in your solution. However, you are asked to completely precipitate the cations. Well — that is a nigh impossible task unless you define a concentration threshold which you want to reach. Your threshold could be $1\,\mathrm{mM}$, $1\,\mathrm{µM}$, $1\,\mathrm{nM}$ or whichever you see fit. Use this desired concentration to recalculate the concentration of chloride ions required for both cations. Then continue on.

I recalculated the value in the spoiler tag for $1\,\mathrm{µM}$ residual cation concentration.


   $4.12\,\mathrm{M}$ chloride, equivalent to $0.687\,\mathrm{mL}$ or 18 drops.
Backwards calculation tells me, that you had a residual lead concentration of ca. $9.58\,\mathrm{µM}$, explaining your second result.