Reactions of phloroglucinol
In aromatic compounds, the enol form is more stable than the keto form due to the aromatic character. In normal conditions, aromatic compounds exist in enol form but in basic conditions, it tautomerises to keto form.  Phloroglucinol predominantly exist in enol form under normal condition. 


iodoform test


Compounds containing ketomethyl group($\ce{-COCH3}$) will give iodoform test. e.g.- propanone, acetophenone etc. Thus phloroglucinol will not give iodoform test because it does not contain ketomethyl group.


ceric ammonium nitrate


Generally, CAN test is done for alcohols which give pink or red colour. But for phenols and phenolic compounds, it gives brown or black test. So, this test helps to differentiate phenols from alcohols. See the detailed question on Chem.SE


2,4 DNP test


Phloroglucinol predominantly exist in enol form under normal condition and thus not give  2,4 DNP test. However, in basic condition (pH- 11-13) it forms 3,5-dihydroxy-2,5-cyclohexadienone dianion,  the deprotonated intermediate in the interconversion of the two forms, referred to as the enolate anion which might give the 2,4 -DNP test. See here

One final note 

Don't take tautomerism into account while experimentally determining the functional group in organic compounds . If you consider tautomerism, there will be two functional group for the compound and you will get confused as of which functional group should be considered. For ex- propanone tautomerises to form prop-1-en-2-ol which contains an alcoholic group and then you will get confused which test should be used- Ester or CAN test(for alcohol) or 2,4 DNP test (for ketone).