What is the hybridization of the C_2 and C_4 carbon atoms,...

According to IUPAC nomenclature rules, the principal functional group determines the numbering of the carbon chain. The ketone group ($>C=O$) has a higher priority than the alkyne group ($-C \equiv C-$).

Numbering the chain to give the ketone the lowest possible locant, we start from left to right:
$C_1$: $\mathrm{CH}_3$
$C_2$: $\mathrm{C}(=\mathrm{O})$
$C_3$: $\mathrm{CH}_2$
$C_4$: $\mathrm{C}$
$C_5$: $\mathrm{CH}$

The $C_2$ atom is bonded to three atoms (one oxygen and two carbons) with no lone pairs, forming $3$ $\sigma$ bonds and $1$ $\pi$ bond. Therefore, its hybridization is $sp^2$.

The $C_4$ atom is bonded to two atoms (two carbons) with no lone pairs, forming $2$ $\sigma$ bonds and $2$ $\pi$ bonds. Therefore, its hybridization is $sp$.

Thus, the hybridization of $C_2$ and $C_4$ are $sp^2$ and $sp$, respectively.

Answer: $sp^2$ and $sp$