Formaldehyde can be distinguished from acetaldehyde by the use of:

Formaldehyde is $\mathrm{HCHO}$ and acetaldehyde is $\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$. Since both are aldehydes and hence are easily oxidized by weak oxidizing agents like Tollen’s reagent, Fehling’s solution. Also, both gives positive Schiff's test positive. So, they cannot be distinguished by them.
Iodoform reaction is used to prove the presence of $\mathrm{R}-\mathrm{CO}-\mathrm{C}{\mathrm{H}}_3$ group of carbonyl and $\mathrm{R}-\mathrm{CH}(\mathrm{OH})-\mathrm{C}{\mathrm{H}}_3$ of alcohol. In this reaction oxidation of the carbonyl group and alcohol group takes place. By oxidation, they form a carboxylic acid. And besides this, yellow precipitation of iodoform takes place in this reaction. This yellow precipitation is the indication that the reaction occurred.
Formaldehyde can be distinguished from acetaldehyde by the use of $\mathrm{NaOH}$ and iodine as formaldehye does not show any reaction while acetaldehye shows this reaction.