Cell voltage with different n factors
If I've understood your question properly, you are asking - in the condition $E^\circ_{\ce{M^3+/M^2+}} =\pu{4 V}$, $E^\circ_{\ce{M^3+/M}} =\pu{1 V}$, $E^\circ_{\ce{Y+/Y}} =\pu{3 V}$ - whether $\ce{M^3+}$ is a stronger oxidising agent than $\ce{Y+}$ irrespective of it going to $\ce{M \text{or} M^2+}$.
This would not be true since $\ce{Y+}$ has a higher reducing potential than $\ce{M^3+/M}$ but not $\ce{M^3+/M^2+}$, and so $\ce{M^3+}$ would be a stronger oxidising agent than $\ce{Y+}$ where it converts into $\ce{M^2+}$ but not when $\ce{M^3+}$ converts into $\ce{M}$.

The method $1$ that you suggest is a very specific case that can only be used for a general redox reaction. By this, I mean a reaction as follows for a cell having it cell notation denoted as $[\ce{N(s)|N^{y+}(aq)|| M^{x+}(aq)|M(s)}]$
\begin{array} {rlllc} \require{cancel}
\ce{N^y+ + ye- &amp;-&gt; N } &amp;E^\circ = E_n^\circ &amp;\quad|\times(-x)\\
\ce{ M^x+ +xe- &amp;-&gt; M}  &amp;E^\circ = E_m^\circ &amp;\quad|\times(y)  \\  \hline
\ce{yM^x+ + xN  +\cancel{(xye-)}  &amp;-&gt; yM + xN^y+ + \cancel{(xye-)}} &amp;E^\circ = E_{\text{cell}}^\circ  \\ 
\end{array}
So here, if you apply method $1$, you get $E^\circ_\text{cell} = E^\circ_m -E^\circ_n$
Let's see whether this works using method $2$ (the free energy varient). Now, we know that $\Delta G$ is additive, $\Delta G^\circ =nFE^\circ$ and that when we multiply the total reaction by a value, we also multiply the $\Delta G$ by the same number. Using these properties we find the $\Delta G$ values for the two reactions.
For just one mole of M being reduced , we get
$$\Delta G_{\text{m}}=xFE^\circ_m \tag{1}$$
For just one mole of N, we get
$$\Delta G_{\text{n}}=yFE^\circ_n \tag{2}$$
Now, according to the above cell reaction, the $\Delta G_{\mathrm {cell}} = y(\Delta G_m) + (-x)\Delta G_n$ which would be equal to
\begin{align}
\Delta G_{\text{cell}} &amp;= (-x)yFE^\circ_n + (y)xFE^\circ_m \\
&amp;= xyF(E^\circ_m -E^\circ_n) \tag{3}
\end{align}
Now, the value of $\Delta G_{\text{cell}}$ in terms of $E^\circ_\text{cell}$, we get:
$$\Delta G_{\text{cell}}=xyFE^\circ_\text{cell} \tag{4}$$
Now substituting this value of $\Delta G_\text{cell}$ into equation (3), we get
\begin{align}
\cancel{(xyF)}E^\circ_\text{cell} &amp;= \cancel{(xyF)}(E^\circ_m -E^\circ_n) \\
\implies E^\circ_\text{cell} &amp;= E^\circ_m -E^\circ_n
\end{align}
For a case where you have the same compound disproportionate, this question and its answers should be enough.