Solution

(i) Amines lose a proton to form amide ion. Alcohols lose a proton to form alkoxide ion.
$$\displaystyle RNH_2 \rightarrow RNH^- + H^+ \\ R-OH \rightarrow R-O^- + H^+$$
$$O$$ is more electronegative than N, the negative charge is more easily accomodated in $$\displaystyle RO^-$$ than in $$\displaystyle R-NH^-$$. Hence, amines are less acidic than alcohols of comparable molecular masses.

(ii) In primary amines, $$N$$ atoms have two $$H$$ atoms which results in extensive intermolecular $$H$$ bonding. In tertiary amines, $$N$$ atoms do not have H atoms and hydrogen bonding is not possible.
Hence, primary amines have higher boiling point than tertiary amines.

(iii) Aliphatic amines stronger bases than aromatic amines due to following reasons:
(a) Aromatic amines have resonance due to which lone pair of electrons on $$N$$ atom is delocalized over benzene ring and is less available for protonation.
(b) The stability of aryl amine ions is lower than the stability of alkyl amines. Protonation of aromatic amines is not favoured.