In carboxylic acids, molecules are more strongly associated followed by alcohols.

In tautomerism, $$\alpha - hydrogen$$    must be present in the molecule. Thus, molecule II will not show tautomerism. As at bridge, double bond is highly unstable. Thus, molecule I will also not show tautomerism.
For III molecule,

So only, III molecule will show tautomerism. Thus, correct option is (A).

$$\eqalign{ & \% \,\,{\text{of}}\,\,P = \frac{{62}}{{222}} \times \frac{x}{w} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{62}}{{222}} \times \frac{{2.6}}{{1.6}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 45.38\% \cr} $$

Structure I and structure II are the resonating forms because the position of atoms remains the same and only redistribution of electrons take place.

Electron deficient species act as electrophiles i.e., $$NO_2^ + .$$

Stronger the acid, weaker the conjugate base. Since acid character follows the order
$${H_2}O > HC \equiv CH > N{H_3} > C{H_3} - C{H_3}$$
( Acid character ),
the basic character of their conjugate bases decreases in the reverse order, i.e.,
$$C{H_3}CH_2^ - > NH_2^ - > HC \equiv {C^ - } > O{H^ - }$$
(Basic character)

The enolic form of ethyl acetoacetate has 16 single bonds i.e. $$16\sigma - $$  bonds and 2 double bonds i.e. $$2\sigma - $$  bonds and $$2\pi - $$  bonds.
Hence, the given structure has $$18\,\sigma - $$  bonds and $$2\pi - $$  bonds.

$$B.P.$$  $$ \propto $$  extent of $$H$$  bonding $$ \propto $$  surface area of molecule.

No explanation is given for this question. Let's discuss the answer together.

TIPS/Formulae :
The bond length decreases in the order.
$$\mathop {s{p^3} - s{p^3}}\limits_{{\text{alkane}}} > \mathop {s{p^2} - s{p^2}}\limits_{{\text{alkene}}} > \mathop {sp - sp}\limits_{{\text{alkyne}}} $$
On the basis of the size of the hybrid orbitals, $$sp$$  orbital should form the shortest and $$s{p^3}$$ orbital the longest bond with other atom.