How to determine the potential of a cadmium zinc cell under standard conditions?
I will like to correct you that there is one problem with your question. The correct standard reduction potential is:
\begin{align}
\ce{Cd^2+  + 2e- &amp;&lt;=&gt; Cd}&amp; E^\circ &amp;= -0.403~\pu{V}.
\end{align}
 And another thing is, you have used symbol $E^\circ$ for oxidation potential of $\ce{Zn}$ which should not be done. You can use there just $E = 0.7628~\pu{V}$ or in a better way, you can just write it with it's reduction potential value as:
\begin{align}
\ce{Zn^2+ + 2e- &amp;&lt;=&gt; Zn}&amp; E^\circ &amp;= -0.7628~\pu{V}.
\end{align}
Now coming to your question, for these two reactions,
\begin{align}
\ce{Zn^2+ + 2e- &amp;&lt;=&gt; Zn}&amp; E^\circ &amp;= -0.7628~\pu{V}\tag1\label{Zn}\\
\ce{Cd^2+ + 2e- &amp;&lt;=&gt; Cd}&amp; E^\circ &amp;= -0.403~\pu{V}\tag2\label{Cd}
\end{align}
$E_\mathrm{cell}$ of any cell reaction is given as
$$E^\circ(\text{cell}) = E^\circ(\text{reduced species}) - E^\circ(\text{oxidized species})$$ 
or
$$E^\circ(\text{cell}) = E^\circ(\text{cathode}) - E^\circ(\text{annode}).$$

Here as $E^\circ$ value of reaction $\eqref{Cd}$ is greater (i.e. towards more positive). Hence $\ce{Cd}$  has greater tendency to accept electrons and get reduced. So the $\ce{Zn}$ is going to give electrons and gets oxidized. Now the half cell reaction will be,
\begin{align}
\ce{Zn &amp;&lt;=&gt; Zn^2+ + 2e^-}\tag{at anode}\label{anode}\\
\ce{Cd^2+ + 2e- &amp;&lt;=&gt; Cd}.\tag{at cathode}\label{cathode}
\end{align}

So according to formula 
\begin{align}
E^\circ(\text{cell}) &amp;= E^\circ(\text{cathode}) - E^\circ(\text{annode})\\
E^\circ(\text{cell}) &amp;= -0.403~\pu{V} - (-0.7628~\pu{V}) = 0.3598~\pu{V}\\
\end{align}

The correct answer is $0.3598~\pu{V}$.