Is (Zn(OH)4)2- acidic or alkaline?
Let's first imagine that one of these complexes is placed in neutral (pH 7) water. If the resulting solution possesses a pH of > 7, we can conclude that the anions show an "alkaline character" as you describe. If the resulting pH is &lt; 7, we'll conclude that the material is contributing an "acidic character." 

The way one might go about answering this question without dunking one of these salts in water is by starting with the a more well know complex such as the hexa-aqua complex of each metal and observing the behavior. 

$\ce{AlCl3 + 6H2O -&gt; Al^3+.6H2O + 3Cl^-}$ Chloride becomes a spectator anion

This $\ce{Al}$ water complex ($\ce{Al^3+.6H2O}$) possesses acidic protons, just like any polyprotic acid (eg. phosphoric acid.) Thus, each proton also has an acid dissociation constant or pKa. The first pKa for $\ce{Al^3+.6H2O}$ is approximately 5 and would thus acidify the solution. The pKa's of each of the subsequent protons donated to solution is higher than the last and eventually, no more protons would be donated to the solution.

With this basic understanding out of the way, let's look at the approximate pKa's of each of the neutral hydroxide salts of each of these metals. 

For $\ce{Al(OH)3}$ the pKa is > 7. This means that by adding another hydroxide to the complex to make it a tetrakis hydroxide complex and would result in an even higher pKa and a more alkaline solution. 

$\ce{Pb(OH)2}$ and $\ce{Zn(OH)2}$ are both weakly basic (pKa > 7) and possessing low solubilites. Adding more hydroxide ligands will make each of them more alkaline and should also make each of them more soluble. The zincate anion $\ce{Zn(OH)4^2-}$ is often formed by dissolving Zinc metal in a strongly alkaline solution.$\ce{^1}$ 

I hope this helps a little.

ref 1. http://depts.washington.edu/chem/facilserv/lecturedemo/BrassPenny-UWDept.ofChemistry.html