Tests for carbonyl compounds

Forms a phenylhydrazone, so X is a carbonyl compound.
Negative Tollens' test, so it is a ketone, not an aldehyde.
Negative iodoform test, so it has no $CH_3CO-$ group.
Reduction gives n-pentane, so the chain is unbranched: X is 3-pentanone, $CH_3CH_2COCH_2CH_3$.
$CH_3CH_2COCH_2CH_3 \xrightarrow{Zn-Hg/HCl} CH_3CH_2CH_2CH_2CH_3$

A yellow precipitate of iodoform ($CHI_3$) is given by compounds having the $CH_3-CO-$ group attached to C or H.
Acetophenone has this group: $C_6H_5COCH_3 \xrightarrow{I_2/NaOH} C_6H_5COONa + CHI_3\downarrow$
In methyl acetate and acetamide the $CH_3CO-$ group is attached to O or N, so they do not give the test.
As per the key the answer is acetophenone.

Both compounds are aldehydes, so both respond to Fehling's, Benedict's and Tollens' tests. These tests cannot tell them apart.
The iodoform test is given only by aldehydes and ketones that have the $CH_3-CO-$ group.
$CH_3CHO$ has this group: $CH_3CHO + 3I_2 + 4NaOH \rightarrow CHI_3 + HCOONa + 3NaI + 3H_2O$, giving a yellow precipitate of iodoform.
$C_6H_5CH_2CHO$ has no $CH_3-CO-$ group, so it gives no precipitate.
So the two are distinguished by the iodoform test.