Why do diacidic and triacidic alkalis dissociate in one step?
Polyprotic acids


In most acids, even the strong acids, the bond to the acidic hydrogen atom is covalent (though frequently very polar). Thus, for acids, the more correct description of what happens in solution is ionization - a covalent bond is breaking heterolytically to form ions. The solvent is involved in solvating the proton, which is why we usually write $\ce{H3O+}$ for protons in water and not $\ce{H+}$. 

Polyprotic acids ionize in separate steps because after the first ionization, the new species has a different strength covalent bond to break. Consider your example of sulfuric acid. Sulfuric acid is a strong acid that ionizes readily and (nearly) completely in water:

$$\ce{H2SO4 + H2O -&gt; HSO4- + H3O+}$$

The resulting hydrogen sulfate anion is a different chemical species. Its bonds have different strengths. The second proton is not as easily ionized. In fact, the hydrogen sulfate anion is considered a weak acid. It does not completely ionize in water. An equilibrium is established with an equilibrium constant of $K_a = 1.02 \times 10^{-2}$. This means that at $\mathrm{pH = 2}$, the hydrogen sulfate anion is 50% ionized: half is hydrogen sulfate and half has ben converted to sulfate.

$$\ce{HSO4- + H2O &lt;=&gt; SO4^2- + H3O+}$$

What evidence do we have for this step-wise ionization? We have titrations! If we add small volumes of a solution of a stong base (say $\ce{NaOH}$) of known concentration to a solution of sulfuric acid and monitor the pH, we can see two distinct equivalence points (the inflection points on the graph where the pH change is very large over a small volume of base added) corresponding to the two ionization events. However, because the first ionization of sulfuric acid is complete before the base is added, a titration  curve of sulfuric acid will only show one equivalence point when two equivalents of base are added (though there is a blip at 1 equivalent). 



If we consider maleic acid $(\ce{H2C4H2O4})$, for which both ionizations are "weak" $(K_1=1.26\times 10^{-2},\ K_2=8.51\times 10^{-7})$, we get a different looking titration curve with inflections at 1 and 2 equivalents of base. If the ionization of both protons occurred simultaneously, we would not see a curve like this. 

$$\ce{H2C4H2O4 +H2O &lt;=&gt; HC4H2O4- + H3O+}\\
\ce{HC4H2O4- + H2O &lt;=&gt; C4H2O4^2- + H3O+}$$




  Bases


Bases that are the hydroxide salts of metals are ionic compounds. Though many of them have poor solubility in water, what is dissolved is completely ionized. Thus, for copper (II) hydroxide in water at equilibrium, there is only solid copper (II) hydroxide ad solubilized copper(II) and hydroxide ions:

$$\ce{Cu(OH)2 (S) &lt;=&gt; Cu^2+ (aq) + 2OH- (aq)}$$

A titration curve for copper (II) hydroxide with a strong acid like hydrochloric acid would show only one equivalence point at two equivalents of acid:



All ionic bases will behave similarly.

However, some bases are covalent and ionize by reacting with water:

$$\ce{NH3 + H2O &lt;=&gt; NH4+ + OH-}$$

If there was to be a base of this type that had two ionizable groups, we would see two equivalence points.