Step 1: Identify the two amines and their classes.
$(C_2H_5)_2NH$ is diethylamine: a secondary amine ($2^\circ$, has one N-H bond). $(C_2H_5)_3N$ is triethylamine: a tertiary amine ($3^\circ$, has no N-H bond).
Step 2: Recall Hinsberg's reagent.
Hinsberg's reagent is benzenesulphonyl chloride ($C_6H_5SO_2Cl$). It is the standard reagent to distinguish between primary ($1^\circ$), secondary ($2^\circ$), and tertiary ($3^\circ$) amines based on different reactions.
Step 3: Reaction of secondary amine (diethylamine) with Hinsberg's reagent.
Diethylamine ($2^\circ$) reacts with $C_6H_5SO_2Cl$ to give a sulphonamide that is insoluble in NaOH (precipitate forms): \[ (C_2H_5)_2NH + C_6H_5SO_2Cl \rightarrow (C_2H_5)_2N-SO_2C_6H_5 + HCl \]
Step 4: Reaction of tertiary amine (triethylamine) with Hinsberg's reagent.
Triethylamine ($3^\circ$) has no N-H bond, so it cannot react with $C_6H_5SO_2Cl$ under normal conditions. No precipitate forms; the mixture remains clear.
Step 5: How this distinguishes the two amines.
With Hinsberg's reagent: Secondary amine gives an insoluble sulphonamide precipitate. Tertiary amine gives no reaction (no precipitate). This clear visual difference distinguishes $(C_2H_5)_2NH$ from $(C_2H_5)_3N$.
Step 6: Verify other options are incorrect for this purpose.
$CHCl_3 + KOH$ (carbylamine test) only works for $1^\circ$ amines. $ZnCl_2 + conc. HCl$ (Lucas test) is for alcohols. $NaOH + I_2$ (iodoform test) is for methyl ketones or ethanol, not amines. \[ \boxed{C_6H_5SO_2Cl \text{ (Hinsberg's reagent)}} \]