How to get a precise answer to a textbook precipitation problem?
Ok, let's do the "complete" and "exact" problem. 

First, I'm just going to ignore the $\ce{Mn^{2+}}$. Using a first pass with the simplifying assumptions would be enough to show that MnS will not ppt. 

Second the problem statement notes that:


  This solution is then saturated with H2S. The solubility of $\ce{H2S}$ is given as 0.1 mol/L and assumed independent of the presence of other substances in solution.  


Chemically $\ce{H2S}$ would be bubbled through the solution to saturate it. While the $\ce{H2S}$ was bubbling $\ce{CuS}$ would be ppt'ing. So when the final concentration of $\ce{H2S}$ is 0.1 molar the ppt of $\ce{CuS}$ is complete.

Let:


The final $\ce{[H2S]_f = 0.1\bar0}$ molar
This means that the final concentration of all the sulfur species ($\ce{[H2S]_f, [HS-]_f,}$ and $\ce{[S^{2-}]_f}$) will be a bit more than $0.1\bar0$ molar. 
The initial and final $\ce{[HClO4]} = 3.\bar 0\times10^{-3}$ molar
From $K_{sp}$ for CuS we know that:
$$K_{sp} = [Cu^{2+}]_f[S^{2-}]_f$$ or $$\ce{[S^{2-}]_f} = \dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}}$$
For $\ce{H2S}$ we can multiply $K_{a1}\times K_{a2}$ to get an overall ionization product $$K_t = K_{a1}\times K_{a2} = 1.2\bar0\times10^{-20} = \dfrac{\ce{[H+]_f[HS-]_f}}{\ce{[H2S]_f}}\times\dfrac{\ce{[H+]_f[S^{2-}]_f}}{\ce{[HS-]_f}} = \dfrac{\ce{[H+]_f^2[S^{2-}]_f}}{\ce{[H2S]_f}}$$
also that
$$\ce{[H+]_f} = \sqrt{\dfrac{K_t\ce{[H2S]_f}}{\ce{[S^{2-}]_f}}} = \sqrt{\dfrac{K_t\ce{[H2S]_f[Cu^{2+}]_f}}{K_{sp}}}$$
and
$$\ce{[HS-]_f} = \dfrac{K_{a1}\ce{[H2S]_f}}{\ce{[H+]_f}} = K_{a1}\ce{[H2S]_f}\sqrt{\dfrac{\ce{[S^{2-}]_f}}{K_t\ce{[H2S]_f}}}=K_{a1}\sqrt{\dfrac{\ce{[H2S]_f[S^{2-}]_f}}{K_t}}= K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}}$$
$\ce{[Cu^{2+}]_f}$ be the final concentration of copper. 
The initial $\ce{[Cu^{2+}]_i}=2\times10^{-4}$ molar which we will assume to be $2.\bar0\times10^{-4}$ molar. Thus the amount of copper that is removed from solution is $2.\bar0\times10^{-4} - \ce{[Cu^{2+}]_f}$
For the final $\ce{[H+]}$ we know that 
$$\ce{[H+]_f = [HClO4] + [HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - \ce{[Cu^{2+}]_f}) - (10^{-7} -[OH-])}$$ or $$ 0 = \ce{[HClO4] + [HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - [Cu^{2+}]_f) - (10^{-7} -\dfrac{K_w}{\ce{[H+]}}) - [H+]}$$ 
substituting for $\ce{[HClO4]}$ 
$$0 = 3.0000\times10^{-3} + \ce{[HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - [Cu^{2+}]_f) - (10^{-7} -\dfrac{K_w}{\ce{[H+]}}) - [H+]}$$ 
and collecting terms
$$0 = 3.3999\times10^{-3} + \ce{[HS^-]_f + 2[S^{2-}]_f - 2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$ 
substituting for $\ce{[S^{2-}]_f}$ 
$$0 = 3.3999\times10^{-3} + \ce{[HS^-]_f} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$ 
substituting for $\ce{[HS^-]_f}$ 
$$0 = 3.3999\times10^{-3} + K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$ 
and finally substituting for $\ce{[H+]_f}$ 
$$0 = 3.3999\times10^{-3} + K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +K_w\sqrt{\dfrac{K_{sp}}{K_t\ce{[H2S]_f[Cu^{2+}]_f}}} - \sqrt{\dfrac{K_t\ce{[H2S]_f[Cu^{2+}]_f}}{K_{sp}}}}$$


Since $\ce{[H2S]_f = 0.10}$ per problem statement, the only variable left is $\ce{[Cu^{2+}]_f}$. I'll leave it to the reader to solve this mess for $\ce{[Cu^{2+}]_f}$, but to 2 significant figures the answer should be $7.7\cdot10^{-21}$.

The whole point in using significant figures to simply the problem is not to get the "wrong" answer, but to avoid a computational mess that doesn't yield any better answer.