Percentage purity of Ba(MnO4)2 sample that completely reacts with given H2O2?
The question does not demand to solve it using "equivalents," I'm going to attempt to solve the problem in universal way, using moles. As OP correctly suggested, this reaction of $\ce{Ba(MnO4)2}$ and $\ce{H2O2}$ is a redox reaction. Since the reaction has taken place in acidic medium and has completely reacted (assuming observation is made by appearance so assuming it was judged by clear solution), the two half reactions should be:

$$
\begin{align}
\ce{MnO4- + 8H+ + 5e- &amp;-&gt; Mn^2+ + 4H2O} &amp; E^\circ &amp;= \pu{1.507 V} \\ 
\ce{H2O2  &amp;-&gt; O2(g) + 2H+ + 2e-} &amp; E^\circ &amp;= \pu{-0.695 V}
\end{align}
$$

Thus, total redox reaction can be written as:

$$\ce{2MnO4- + 6H+ + 5H2O2 -&gt; 2Mn^2+ + 5O2(g) + 8H2O} \quad E^\circ_\mathrm{rxn} = \pu{0.812 V}$$

The positive $E^\circ_\mathrm{rxn}$ means the reaction is spontaneous. And it also shows that you need $\pu{5 mol}$ of $\ce{H2O2}$ to completely react with $\pu{2 mol}$ of $\ce{MnO4-}$. Since $\pu{1 mol}$ of $\ce{Ba(MnO4)2}$ contains $\pu{2 mol}$ of $\ce{MnO4-}$, it is correct to say that $\pu{5 mol}$ of $\ce{H2O2}$ would completely react with $\pu{1 mol}$ of $\ce{Ba(MnO4)2}$. 

Suppose that $\pu{40 g}$ of impure sample contains $x~\pu{g}$ of $\ce{Ba(MnO4)2}$. Then, amount of of $\ce{Ba(MnO4)2}$ in the sample is

$$\frac{x~\pu{g}}{\pu{375 g\cdot mol-1}} = \frac{x}{\pu{375 mol}}.$$

To completely react with that amount, you need

$$\left(5\cdot\frac{x}{375}\pu{mol}\right)~\text{of}~\ce{H2O2}.$$

Thus,

$$5\times\frac{x}{375}\pu{mol} = \pu{3\frac{mol}{L}}\times\pu{125 mL}\times \pu{10^{-3} \frac{L}{mL}} = 3\times \pu{0.125 mol} \label{eq:1}\tag{1}$$

$$\therefore x = \pu{\frac{ 3\times 0.125 \times 375}{5} g} = \pu{28.1 g}$$

Thus,

$$\text{percentage of $\ce{Ba(MnO4)2}$ in the $\pu{40 g}$ of sample} = \frac{28.1}{40}\times 100 = 70.2$$

Note that, the equation $\eqref{eq:1}$ is exactly the same as yours (with minieq.)