Step 1: Read the clues from the formula.
\(C_5H_{10}O\) has one degree of unsaturation, which fits a C=O group, so X is an aldehyde or a ketone.
Step 2: Use the tests as filters.
Tollen's test is positive only for aldehydes. The iodoform test is positive for a \(CH_3CO-\) (methyl ketone) group. Cannizzaro needs an aldehyde with no alpha hydrogen.
Step 3: Solve part (I).
No Tollen's test means it is a ketone, and a positive iodoform test means it is a methyl ketone. The methyl ketone of \(C_5H_{10}O\) is pentan-2-one, \(CH_3COCH_2CH_2CH_3\).
Step 4: Solve part (II).
No Tollen's and no iodoform, but it does Aldol condensation, so it is a ketone that has alpha hydrogens but is not a methyl ketone. Pentan-3-one, \(CH_3CH_2COCH_2CH_3\), fits well.
Step 5: Solve part (III).
Cannizzaro happens with an aldehyde that has no alpha hydrogen. The aldehyde with the carbonyl carbon attached to no alpha H is 2,2-dimethylpropanal (pivaldehyde), \((CH_3)_3C-CHO\).
Answer: (I) Pentan-2-one, \(CH_3COCH_2CH_2CH_3\); (II) Pentan-3-one, \(CH_3CH_2COCH_2CH_3\); (III) 2,2-dimethylpropanal, \((CH_3)_3C-CHO\).