Determining the order of basicity of acetate, phenoxide, hydroperoxide, and hydroxide ions
You are partly wrong and partly right. The correct order is $4 &gt; 3 &gt; 2 &gt; 1$ ($\mathrm pK_\mathrm a$ data taken from the Evans table):

$$\begin{array}{cc}
\hline
\text{Molecule} &amp; \mathrm{p}K_\mathrm{a} \\
\hline
\ce{CH3COOH} &amp; 4.76 \\
\ce{C6H5OH} &amp; 9.95 \\
\ce{H2O2} &amp; 11.6 \\
\ce{H2O} &amp; 15.7 \\
\hline
\end{array}$$

Between acetate and phenoxide, phenoxide does have more (non-trivial) resonance forms, but all of them involve putting a negative charge on carbon which is hardly favourable. So, the stabilisation derived from resonance is not much at all.



On the other hand, in acetate, the other major resonance form has the negative charge on an electronegative oxygen. The moral of the story is that the actual number of resonance contributors is not the only deciding factor:



Between hydroperoxide and hydroxide, the $\mathrm{p}K_\mathrm{a}$ values show that you are right in saying that the second oxygen in the hydroperoxide ion stabilises the negative charge. This operates entirely via an inductive effect, though. There is no resonance involved, which partly explains why both ions are more basic than phenoxide and acetate.