Balancing and combining the half equations in redox
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.



$$\ce{H2 + NO -&gt; NH3 + H2O}$$

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.*
$$\small\ce{H2 + NO -&gt; NH3 + H+ + OH-}$$
*-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 $\ce{H2 -&gt; NH3}$  (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 $\ce{H2 -&gt; H2O}$.

$$
\begin{array}{l|r}
\text{Oxidation Half} &amp; \text{Reduction half} \\
\hline
\ce{H2 -&gt; H+ + OH-} &amp; \ce{NO -&gt; NH3}
\end{array}
$$

Step 3: Balance only those atoms undergoing redox

$$
\begin{array}{l|r}
\text{Oxidation Half} &amp; \text{Reduction half} \\
\hline
\ce{H2 -&gt; H+ + OH-} &amp; \ce{NO -&gt; NH3} \\
\text{(because Hydrogen is } &amp; \text{(because only Nitrogen } \\
\text{undergoing redox but} &amp; \text{is undergoing redox,} \\
\text{Oxygen is not.)} &amp; \text{it's already balanced)} \\
\end{array}
$$

Step 4: Balance Oxygen by adding water($\ \small\ce{H2O}\ $)

$$
\begin{array}{l|r}
\text{Oxidation Half} &amp; \text{Reduction half} \\
\hline
\ce{H2 + H2O -&gt; H+ + OH-} &amp; \ce{NO -&gt; NH3 + H2O}
\end{array}
$$

Step 5a: Balance Hydrogen by adding $\ \small\ce{H+}$

$$
\begin{array}{l|r}
\text{Oxidation Half} &amp; \text{Reduction half} \\
\hline
\ce{H2 + H2O -&gt; H+ + OH- + 2H+} &amp; \ce{NO + 5H+ -&gt; NH3 + H2O} \\
\ce{H2 + H2O -&gt; [H+ + OH- ] + 2H+} &amp; \ce{NO + 5H+ -&gt; NH3 + H2O} \\
\ce{H2 + H2O -&gt; \qquad H2O \qquad + 2H+} &amp; \ce{NO + 5H+ -&gt; NH3 + H2O} \\
\hline
\ce{H2 + \qquad-&gt; \qquad \qquad \qquad \quad 2H+} &amp; \ce{NO + 5H+ -&gt; NH3 + H2O}\\
\hline
\end{array}
$$

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.
$$
\begin{array}{l|r}
\text{Oxidation Half} &amp; \text{Reduction half} \\
\hline
\ce{H2 \qquad -&gt; 2H+} &amp; \ce{NO + 5H+ \qquad \qquad -&gt; NH3 + H2O \qquad \qquad} \\
\ce{H2 + 2OH- -&gt; 2H+ + 2OH-} &amp; \ce{NO + 5H+ + 5OH- -&gt; NH3 + H2O + 5OH-} \\
\ce{H2 + 2OH- -&gt; [2H+ + 2OH- ]} &amp; \ce{NO + [5H+ + 5OH- ]-&gt; NH3 + H2O + 5OH-} \\
\ce{H2 + 2OH- -&gt; \quad \quad 2H2O} &amp; \ce{NO + \qquad 5H2O \qquad -&gt; NH3 + H2O + 5OH-} \\
\hline
\ce{H2 + 2OH- -&gt; \quad \quad 2H2O} &amp; \ce{NO + \qquad 4H2O \qquad -&gt; NH3 + 5OH- \quad \qquad  }\\
\hline
\end{array}
$$

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

$$
\begin{array}{l|r}
\text{Oxidation Half} &amp; \text{Reduction half}  \\
\hline
(acidic)\quad \ce{H2  -&gt; 2H+ + 2e-} &amp; \ce{NO + 5H+ + 5e- -&gt; NH3 + H2O} \\
\hline
(basic) \quad \ce{H2 + 2OH- -&gt; 2H2O + 2e-} &amp; \ce{NO + 4H2O + 5e- -&gt; NH3 + 5OH-} \\
\hline
\end{array}
$$

&rarr; Verify correctness by ensuring electrons are on right hand side in oxidation half and on left hand side in reduction half.

&rarr; 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.)

$$
\begin{array}{l|r}
\text{Oxidation Half} &amp; \text{Reduction half} \\
\hline
(acidic) \ce{5H2 -&gt; 10H+ + 10e-} &amp; \ce{2NO + 10H+ + 10e- -&gt; 2NH3 + 2H2O}\\
\hline
(basic) \ce{5H2 + 10OH- -&gt; 10H2O + 10e-} &amp; \ce{2NO + 8H2O + 10e- -&gt; 2NH3 + 10OH-} \\
\hline
\end{array}
$$



Finally add the two halves and cancel out the common cmopounds and you get the balanced equation
Acidic Medium
$$\begin{matrix}
\ce{&amp;5H2 + &amp;2NO + &amp;10H+ + &amp;10e- &amp;-&gt; &amp;10H+ + &amp;10e- + &amp;2NH3 + &amp;2H2O} \\
\ce{&amp;5H2 + &amp;2NO &amp; &amp; &amp;-&gt; &amp; &amp; &amp;2NH3 + &amp;2H2O}\\
\end{matrix}$$

Basic Medium
$$\begin{matrix}
\ce{&amp;5H2 + &amp;10OH- + &amp;2NO + &amp;8H2O + &amp;10e- &amp;-&gt; &amp;10H2O + &amp;10e- + &amp;2NH3- + &amp;10OH-} \\
\ce{&amp;5H2 + &amp; &amp;2NO &amp; &amp; &amp;-&gt; &amp;2H2O + &amp; &amp;2NH3}\\
\end{matrix}$$



In this case the oxidation half could have been directly/trivially stated as $\small\ce{H2 -&gt; 2H+}$

You can also carry on $\small\ce{H+}$ ion till the end net reaction and then perform Step 5b which is less tedious.