It depends on the structure of the ether. Ethers connected to tertiary carbon will fragment via SN1, and ethers connected to primary carbons will fragment via SN2. It’s the middle ground (secondary) where there will be a mixture of the two pathways.

It is difficult to predict with unsymmetrical di-alkyl ethers. What often happens is that excess H-I is used such that both alkyl fragments are converted into alkyl iodides.

It is easier with aryl alkyl ethers, which will never result in cleavage of the aryl-O bond

Hard to predict from first principles without doing the experiment.
However, one way to know for sure is to use a large excess of HI, which will give you both isopropyl iodide and methyl iodide.

Correct.