Assuming this reaction is taking place in aqueous phase, you can follow the conventional 7 steps for balancing redox reactions although this one does require some extra thoughts.
I will be writing the equation as it should be after each step.
$$\ce{H2 + NO -> NH3 + H2O}$$
Step 1: Ionise the required compounds and remove spectator ions
None of the compounds on either side are ionic other than $\small\ce{H2O}$ also we have no spectator ions as well.*
$$\small\ce{H2 + NO -> NH3 + H+ + OH-}$$
*-Spectator ions are ions which don't participate in the reaction or retain their oxidation state.
Step 2: Split into Oxidation and reduction halves
This is where the thought is required, if you decide of taking the oxidation half as $\ce{H2 -> NH3}$ (you can't simply take $\ce{H3^3-}$ as $\ce{NH3}$ is not ionic rather covalent) you would never be able to balance the nitrogen as the only source of nitrogen is $\ce{NO}$ if you observe the left hand side. Also, since you know that Nitrogen is being reduced (+2 -> -3) the other element, Hydrogen must get oxidised (0 -> +1).
So it is decided that hydrogen cannot convert into ammonia, leaving us with only one option $\ce{H2 -> H2O}$.
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 -> H+ + OH-} & \ce{NO -> NH3}
\end{array}
$$
Step 3: Balance only those atoms undergoing redox
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 -> H+ + OH-} & \ce{NO -> NH3} \\
\text{(because Hydrogen is } & \text{(because only Nitrogen } \\
\text{undergoing redox but} & \text{is undergoing redox,} \\
\text{Oxygen is not.)} & \text{it's already balanced)} \\
\end{array}
$$
Step 4: Balance Oxygen by adding water($\ \small\ce{H2O}\ $)
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 + H2O -> H+ + OH-} & \ce{NO -> NH3 + H2O}
\end{array}
$$
Step 5a: Balance Hydrogen by adding $\ \small\ce{H+}$
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 + H2O -> H+ + OH- + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\
\ce{H2 + H2O -> [H+ + OH- ] + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\
\ce{H2 + H2O -> \qquad H2O \qquad + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\
\hline
\ce{H2 + \qquad-> \qquad \qquad \qquad \quad 2H+} & \ce{NO + 5H+ -> NH3 + H2O}\\
\hline
\end{array}
$$
Step 5b: (Only in case of basic medium) Adding $\ \small\ce{OH-}$ to $\ \small\ce{H+}$
This step must follow Step 5a and means nothing on it's own.
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 \qquad -> 2H+} & \ce{NO + 5H+ \qquad \qquad -> NH3 + H2O \qquad \qquad} \\
\ce{H2 + 2OH- -> 2H+ + 2OH-} & \ce{NO + 5H+ + 5OH- -> NH3 + H2O + 5OH-} \\
\ce{H2 + 2OH- -> [2H+ + 2OH- ]} & \ce{NO + [5H+ + 5OH- ]-> NH3 + H2O + 5OH-} \\
\ce{H2 + 2OH- -> \quad \quad 2H2O} & \ce{NO + \qquad 5H2O \qquad -> NH3 + H2O + 5OH-} \\
\hline
\ce{H2 + 2OH- -> \quad \quad 2H2O} & \ce{NO + \qquad 4H2O \qquad -> NH3 + 5OH- \quad \qquad }\\
\hline
\end{array}
$$
The medium being acidic or basic merely determines whether $\ \small\ce{H+}$ or $\ \small\ce{OH-}$ will be seen in the final redox equation.
Till the above steps mass has been balanced, now charge balancing.
Step 6: Balance charge in each side of each half by adding electrons
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
(acidic)\quad \ce{H2 -> 2H+ + 2e-} & \ce{NO + 5H+ + 5e- -> NH3 + H2O} \\
\hline
(basic) \quad \ce{H2 + 2OH- -> 2H2O + 2e-} & \ce{NO + 4H2O + 5e- -> NH3 + 5OH-} \\
\hline
\end{array}
$$
→ Verify correctness by ensuring electrons are on right hand side in oxidation half and on left hand side in reduction half.
→ Also verify if net change in oxidation state is equal to the number of electrons appearing in each half.
eg. Hydrogen is going from 0 to +1 oxidation state and there are 2 Hydrogen atoms in the left side of the oxidation half undergoing oxidation(don't count $\small\ce{H}$ from $\small\ce{OH-}$), meaning a net change of '2', which is equal to the number of electrons.
Similarly Nitrogen is going from +2 to -3 and there is just one Nitrogen atom in the left side of the reduction half meaning a net change of '5' which is equal to the number of electrons appearing in the equation. (This is what goes on to be later called as the 'n-factor' or the 'valency-factor')
Step 7: Make 'net electrons produced' equal to 'net electrons consumed'
Multiply oxidation half by 5 and reduction half by 2 (this is basically like taking the LCM and multiplying both halves to make the number of electrons equal to the LCM, here 10.)
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
(acidic) \ce{5H2 -> 10H+ + 10e-} & \ce{2NO + 10H+ + 10e- -> 2NH3 + 2H2O}\\
\hline
(basic) \ce{5H2 + 10OH- -> 10H2O + 10e-} & \ce{2NO + 8H2O + 10e- -> 2NH3 + 10OH-} \\
\hline
\end{array}
$$
Finally add the two halves and cancel out the common cmopounds and you get the balanced equation
Acidic Medium
$$\begin{matrix}
\ce{&5H2 + &2NO + &10H+ + &10e- &-> &10H+ + &10e- + &2NH3 + &2H2O} \\
\ce{&5H2 + &2NO & & &-> & & &2NH3 + &2H2O}\\
\end{matrix}$$
Basic Medium
$$\begin{matrix}
\ce{&5H2 + &10OH- + &2NO + &8H2O + &10e- &-> &10H2O + &10e- + &2NH3- + &10OH-} \\
\ce{&5H2 + & &2NO & & &-> &2H2O + & &2NH3}\\
\end{matrix}$$
In this case the oxidation half could have been directly/trivially stated as $\small\ce{H2 -> 2H+}$
You can also carry on $\small\ce{H+}$ ion till the end net reaction and then perform Step 5b which is less tedious.
