Isn't water a Lewis base in the ammonia–water reaction?
Your reasoning is correct. Perhaps the answer key is mistaken.
For part (a), in the reaction
$$\ce{NH3 + H2O &lt;=&gt; NH4+ + OH-}$$
water is donating a proton $(\ce{H+}$ ion) and hence is behaving as a Brønsted acid. Since all Brønsted  acids are Lewis acids, water is behaving as a Lewis acid.
For part (d) as well, the half reactions are:
$$
\begin{align}
\ce{2 NH3 &amp;-&gt; 2 NH2- + 2 H+} \\
\ce{2 Na &amp;-&gt; 2 Na+ + 2 e-}
\end{align}
$$
It's clear that $\ce{NH3}$ is acting as an acid and $\ce{Na}$ is acting as a Lewis base by donating electrons. The $\ce{NH2-}$ ion pairs up with the $\ce{Na+}$ ion to form $\ce{NaNH2}$ and hydrogen accepts the electrons and is liberated as $\ce{H2}.$
EDIT: As pointed out by Mithoron in the comments, the reduction of ammonia is not so simple. Sodium dissolves in ammonia to give a deep-blue solution containing ammoniated electrons (reference here):
$$\ce{Na + \text{(x+y)}NH3 &lt;=&gt; Na(NH3)_x^+ + e(NH3)_y^-}$$
The $\ce{e(NH3)_y^-}$ eventually loses a hydrogen atom to give $\ce{NH2-}$, which then combines with the $\ce{Na+}$ giving the product.