Determining the redox half-reactions of single displacement reactions
Let's take the first equation in the question by example:
$$\ce{Mn(OH)2 (s) + H2O2 (aq) -&gt; Mn(OH)3- (s)}$$
In this equation ion $\ce{Mn(II)}$ is oxidised into $\ce{Mn(III)}$, and oxygen in $\ce{H2O2}$ (oxidation number -1) is reduced into ion hydroxide (oxidation number -2):
$$
\begin{align}
\ce{Mn^{2+} &amp;-&gt; Mn^{3+} +e^- \tag 1}\\
\ce{H_2O2 (aq) + 2e^- &amp;-&gt; 2OH^- \tag 2}
\end{align}
$$ 
If we multiply the first equation by 2 and add the equations $(1)$ and $(2)$: 
$$\phantom{iiii}\ce{2Mn^{2+} + H_2O2 (aq) -&gt; 2 Mn^{3+} + 2OH^- \tag 3}$$ 
By adding four ions hydroxide to the two sides of equation $(3)$: 
$$\ce{2Mn^{2+} + H_2O2 (aq) +4OH^- -&gt; Mn^{3+} + 6OH^- }\phantom{iiiiiiiiii}$$ 
By rearranging the equation $(3)$, we find:
$$\ce{2Mn(OH)2 (s) + H2O2 (aq) -&gt; 2 Mn(OH)3 (s) \tag 4}\phantom{iiiiiii}$$
We do the same for the other equations.