Solution

$$\displaystyle Li$$ and $$\displaystyle NH_3$$
The metal ‘M’ forms nitride $$\displaystyle M_3N$$. Since N has valency 3, metal ‘M’ has valence 1. Hence, ‘M’ can be either $$\displaystyle Li$$ or $$\displaystyle Na$$. So we can rule out $$\displaystyle Ba$$ and $$\displaystyle Al$$.

$$\displaystyle M_3N$$ reacts with water to produce gas ‘B’. This is given by $$\displaystyle Li$$
$$\displaystyle Li_3N +3H_2O \rightarrow 3LiOH + NH_3$$
Thus, gas ‘B’ is $$\displaystyle ammonia$$

$$\displaystyle Ammonia$$ reacts with aqueous $$\displaystyle CuSO_4$$ solution to form deep blue compound $$\displaystyle [Cu(NH_3)_4]SO_4$$
$$\displaystyle NH_3+CuSO_4 \rightarrow [Cu(NH_3)_4]SO_4$$