Balancing redox reaction of zinc and hot diluted nitric acid
The oxidation state of $\ce{Zn}$ as calculated by you is incorrect. The correct oxidation state of $\ce{Zn}$ is $+2$ as it is associated with two mono negatively charged $\ce{NO3-}$ ions.
Thus the unbalanced reaction with correct oxidation states would be:
$$\ce{\overset{0}{Zn} + \overset{+1}{H}\overset{+5}{N}\overset{-2}{O_3}\longrightarrow \overset{+2}{Zn}(\overset{+5}{N}\overset{-2}{O_3})2 +\overset{-3}{N}\overset{+1}{H_4}\overset{+5}{N}\overset{-2}{O_3} +\overset{+1}{H2}\overset{-2}{O}}
$$
Now the half-cell reactions are as follows:
$$
\ce{Zn -&gt; Zn^{2+} + 2e-}\\
\ce{NO3- + 10H+ + 8e- -&gt; NH4+ + 3H2O}\\
$$
Thus the balanced iconic reaction would be:
$$\ce{4Zn + NO3- + 10H+ -&gt; 4Zn^{2+} + NH4+ + 3H2O}\\\\$$
Now adding nine $\ce{NO3-}$ spectator ions to the ionic reaction gives us the standard molecular reaction.
$$\ce{4Zn + 10HNO3 -&gt; 4Zn(NO3)2 + NH4NO3 + 3H2O}\\\\$$
What did you learn from this?
Always try to write the half-cell reaction in ionic form and only involve the ions which undergo oxidation state change. Don't bother yourself with the spectator ions at first. After you balance the equation in an ionic  form adjust it to the molecular form by adding relevantly needed ions. Identify the oxidizing and the reducing species is the key.