Why does Cu(OH)2, which is generally insoluble due to low Ksp become soluble in excess of a base, such as NaOH?
Also, can the equilibrium $$\ce{Cu^{2+} + 6H2O &lt;=&gt; [Cu(H2O)6]^{2+}}$$ participate in the solubility process in any case?


This equilibrium does contribute to the solubility of copper(II), but probably not in the way it appears you are envisioning it to.  When copper(II) is fully dissolved in moderately acidic solution, as long as the anion of the acid is a weak ligand for copper (e.g., $\ce{ClO4-}$ or $\ce{NO3-}$), the copper ions are dissolved as the aquo complex, $\ce{[Cu(H2O)6]^{2+}}$.

As the $\mathrm{pH}$ is adjusted upward with, e.g., $\ce{NaOH}$, though, other reactions come into play (see below) and the hexaaquo species contributes negligibly to the overall solubility behavior.




  Where does the "excess" criteria come into the picture?


To answer this question, one really must first understand what's actually going on when solvated metal ions undergo acid/base reactions.  As I indicated above, in acidic solution copper(II) exists as an aquo complex, typically written as the hexaaquo species, $\ce{[Cu(H2O)6]^{2+}}$.  As the $\mathrm{pH}$ rises, the following series of acid-base ("hydrolysis") reactions occurs$^\ddagger$:

$$
\ce{[Cu(H2O)6]^2+ + OH- &lt;=&gt; [Cu(H2O)5(OH)]+ + H2O}\tag{1}
$$

$$
\ce{[Cu(H2O)5(OH)]+ + OH- &lt;=&gt; [Cu(H2O)4(OH)2]^0 + H2O}\tag{2}
$$

$$
\ce{[Cu(H2O)4(OH)2]^0 + OH- &lt;=&gt; [Cu(H2O)3(OH)3]- + H2O}\tag{3}
$$

$$
\ce{[Cu(H2O)3(OH)3]- + OH- &lt;=&gt; [Cu(H2O)2(OH)4]^2- + H2O}\tag{4}
$$

(For simplicity, I have omitted the $\ce{M2L2}$ hydrolysis complex reported in Vuceta$^\ddagger$, and have written the above in terms of addition of alkali so as to highlight the similarities among the series.)

So, to some extent, the copper can be treated much like a polyprotic acid: as hydroxide ion is added, the copper is able to "absorb" it, forming water/hydroxide complexes with increasing numbers of hydroxides attached.

Per the low $K_\mathrm{sp}$ of $\ce{Cu(OH)2}$, however, the following (fictitious) solubility relation must also be considered:

$$
\ce{ [Cu(H2O)4(OH)2]^0 (aq) &lt;=&gt; Cu(OH)2 (s) + 4H2O} \tag{5}
$$

Eq. $(5)$ is a re-interpretation of the typical solubility expression:

$$
\ce{[Cu^2+][OH^{-}]^2} \leq K_\mathrm{sp} \tag{6}
$$

Qualitatively, Eq. $(5)$/Eq. $(6)$ indicates that the copper existing as $\ce{[Cu(H2O)4(OH)2]^0}$ is limited to a certain concentration in solution, depending on the conditions of the system.  

Thus, as base is added to an acidic solution of copper and the equilibria are swung to a range that favors $\ce{[Cu(H2O)4(OH)2]^0}$/$\ce{Cu(OH)2}$, much of the copper precipitates out.  But, as more base is added, eventually there's enough hydroxide present that the equilibria can start to shift toward $\ce{[Cu(H2O)3(OH)3]-}$ and $\ce{[Cu(H2O)2(OH)4]^2-}$, and the copper will start to redissolve.



This, finally, leads back to question: what does "excess" mean in this context?

To help answer this question, I input the stability constants from Vuceta$^\ddagger$ into HySS, a program that is designed to simulate these sorts of equilibria and that is able to handle precipitates like $\ce{Cu(OH)2 (s)}$.  The HySS file used to generate the below figures can be downloaded from Google Drive here.  The below figure shows the simulated speciation diagram for $10~\mu\mathrm M$ total copper over a $\mathrm{pH}$ range of $4$ to $14$ (all ligating water molecules are omitted from the labels; click the image for a larger version):

&nbsp;&nbsp;&nbsp;&nbsp;

As expected, the copper is fully dissolved at an acidic $\mathrm{pH}$ of $4$, but starts to precipitate at around $\mathrm{pH}~7$.  Coincidentally, the concentration of the dissolved $\ce{[Cu(H2O)4(OH)2]^0}$ is predicted to be equal to the "concentration" of the precipitated $\ce{Cu(OH)2 (s)}$ (mass of solid divided by the liquid volume) over much of the insoluble range. Starting at about $\mathrm{pH}~11.5$, the base is in sufficient excess (~$316$-fold) over the copper to begin to re-dissolve it.  Only around $\mathrm{pH}~13$ or so does all of the solid finally re-dissolve, at a ~$10,000$-fold excess of base.

At the higher copper concentration of $10~\mathrm{mM}$, re-dissolution has hardly begun to happen at all by $\mathrm{pH~14}$ (click image for larger version):

&nbsp;&nbsp;&nbsp;&nbsp;

This is perfectly sensible, as the base is only in excess of the copper by ~$100$-fold at that $\mathrm{pH}$.  In order to observe re-dissolution at this copper concentration, one would presumably have to use extraordinarily concentrated alkali in order to achieve a sufficient excess of base.



$^\ddagger$ Vuceta &amp; Morgan, Limnology and Oceanography 22(4): 742, 1977, doi:10.4319/lo.1977.22.4.0742