The important thing to note with electrochemistry questions like these is that, if you add two equations together to get a third, their $\Delta G^\circ$ can be added, but their $E^\circ$ cannot.
Your half-equations are incomplete, which might not be the worst thing, but it certainly doesn't help your case. Using $\Delta G^\circ = -nFE^\circ$, we have the two equations
$$\begin{align}
\ce{MnO4- + 8H+ + 5e-} &\ce{-> Mn^2+ + 4H2O} & \Delta G_1^\circ &= -728.46~\mathrm{kJ~mol^{-1}} \tag{1} \\
\ce{MnO2 + 4H+ + 2e-} &\ce{-> Mn^2+ + 2H2O} & \Delta G_2^\circ &= -237.35~\mathrm{kJ~mol^{-1}} \tag{2}
\end{align}$$
The trick is to find the appropriate combination of equations $(1)$ and $(2)$ that give the equation $(3)$, which you are interested in:
$$\ce{MnO4- + 4H+ + 3e- -> MnO2 + 2H2O} \qquad \qquad \Delta G_3^\circ = \,\,? \qquad \qquad \tag{3}$$
If you just look at the manganese-containing species, it's actually not too hard to see that $(1) - (2) = (3)$. That's why I said it's probably not the worst thing that you omitted $\ce{H2O}$ in your half-equations - since it doesn't really make a difference in terms of finding the answer - but it's just not good practice to write something that's incorrect. Anyway, that means that:
$$\begin{align}
\Delta G_3^\circ &= \Delta G_1^\circ - \Delta G_2^\circ \\
&= -728.46~\mathrm{kJ~mol^{-1}} + 237.35~\mathrm{kJ~mol^{-1}} \\
&= 491.11~\mathrm{kJ~mol^{-1}} \\
E_3^\circ &= \frac{-\Delta G_3^\circ}{3F} \\
&= +1.697~\mathrm{V}
\end{align}$$
If you obtain a slightly different answer, it may be due to rounding inconsistencies, or perhaps because I used more accurate values for $F$ and the initially given reduction potentials. In any case, you should get an answer around $+1.70~\mathrm{V}$.
