How to calculate the potential via Hess's law? Is my textbook wrong about this?
You have to take into account, that the given reactions are only reductions, while you need to have an oxidation, too. So one of the equations has to be reversed.

\begin{aligned}
(A)E^\circ = -0.10~\mathrm{V}&amp;&amp;\ce{CO3^{2-} + 7H2O + 8e- &amp;-&gt; CH4 + 10 {}^{-}OH}  \\
(B)E^\circ = +0.55~\mathrm{V}&amp;&amp;\ce{O2 + 2H2O + 4e- &amp;-&gt; 4 {}^{-}OH}  \\\hline
\color{\red}{-1}\cdot(A) + 2\cdot(B)
   &amp;&amp;\ce{CH4 + 2O2 + 10{}^{-}OH + 4H2O&amp;-&gt; CO3^{2-} + 7H2O + 4 {}^{-}OH}\\
\implies
   &amp;&amp;\ce{CH4 + 2O2 + 2{}^{-}OH &amp;-&gt; CO3^{2-} + 3H2O}
\end{aligned}

Equation $(A)$ gives you the reduction of the carbonate ions, hence the oxidation of methane. Equation $(B)$ is the reduction of oxygen. You have a galvanic cell of the form $\ce{CH4|CO3^{2-}||O2|{}^{-}OH}$, hence the standard potential difference is
\begin{aligned}
\Delta E^\circ(\ce{CH4|CO3^{2-}||O2|{}^{-}OH})~
               &amp;= E^\circ(\ce{O2|{}^{-}OH}) -E^\circ(\ce{CH4|CO3^{2-}})\\
               &amp;= E^\circ(\ce{O2|{}^{-}OH}) 
                  -\left[\color{\red}{-1}\cdot E^\circ(\ce{CO3^{2-}|CH4})\right]\\
               &amp;= +0.55~\mathrm{V}          
                  -\left[\color{\red}{-1}\cdot (-0.10~\mathrm{V})\right]\\
\Delta E^\circ(\ce{CH4|CO3^{2-}||O2|{}^{-}OH})~
               &amp;= +0.45~\mathrm{V}
\end{aligned}



Just to address the second edit of the OP and to clarify:
\begin{aligned}
 (1): &amp;&amp;\ce{3 Ag+(aq) + Fe (s)   &amp;-&gt; 3 Ag (s) + Fe^{3+} (aq)}\\
 (2): &amp;&amp;\ce{3 Ag+(aq) + 3 e-     &amp;-&gt; 3 Ag (s)} &amp; E^\circ &amp;= 0.80~\mathrm{V}\\
 (3): &amp;&amp;\ce{Fe^{3+} (aq) + 3 e- &amp;-&gt; 3 Fe (s)} &amp; E^\circ &amp;= -0.04~\mathrm{V}\\
\end{aligned}

The sign of the second equation has to be flipped and then subsequently subtracted from the first equation, hence the potential that should be found is $\Delta E^\circ = 0.74~\mathrm{V}$. Therefore, I believe your textbook is wrong.