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- &-> N } &E^\circ = E_n^\circ &\quad|\times(-x)\\
\ce{ M^x+ +xe- &-> M}  &E^\circ = E_m^\circ &\quad|\times(y)  \\  \hline
\ce{yM^x+ + xN  +\cancel{(xye-)}  &-> yM + xN^y+ + \cancel{(xye-)}} &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}} &= (-x)yFE^\circ_n + (y)xFE^\circ_m \\
&= 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} &= \cancel{(xyF)}(E^\circ_m -E^\circ_n) \\
\implies E^\circ_\text{cell} &= E^\circ_m -E^\circ_n
\end{align}

For a case where you have the same compound disproportionate, [this question and its answers](https://chemistry.stackexchange.com/q/136774/95133) should be enough.