I would like to elaborate more on Ron's answer to make it easier to understand. In particular, I would like to elaborate on the significance of acidity on the reaction mechanism.
Ron mentions that:
Acidic conditions favour the $\mathrm{S_N}1$ mechanism while non-acidic conditions favour the $\mathrm{S_N}2$ mechanism.
Acidic conditions
Initially, I had a lot of problems understanding this too. Upon discussion with my friend Ian, I was able to understand this fully. I was puzzled as to why there was a preference for one mechanism over the other in the different conditions. Let me explain why this preference exists in detail.
When using the reagent $\ce{HI}$, the conditions are acidic, meaning that there is an abundance of proton-donating species in the solution. Under such conditions, the oxygen atom in the ether is protonated. This gives the oxygen atom a formal positive charge, causing it to be more electron-withdrawing, resulting in larger partial positive charges on the two carbon atoms bonded to it. This also provides impetus for the breaking of the $\ce{C-O}$ bond to break. Now, we have two options, the $\ce{C-O}$ bond to the t-butyl group or that to the ethyl group. Since this bond breaks to form a carbocation, it necessarily means that a more stable carbocation would be preferred. Thus, this $\ce{C-O}$ bond would break to form the t-butyl carbocation. Eventually, the products ethanol and 2-iodo-2-methylpropane are formed.
This reaction follows the $\mathrm{S_N}1$ mechanism due to the protonation event which allowed the $\ce{C-O}$ bond to break more easily.
Non-acidic conditions
When using the reagent $\ce{KI}$ in non-acidic condiitions, there is no abundance of proton-donating species in solution. Under such conditions, the oxygen atom in the ether is not protonated significantly. Thus, the impetus to break the $\ce{C-O}$ bond to form the carbocation is not present. The mechanism of the nucleophilic substitution reaction would thus be a concerted one (i.e. $\mathrm{S_N}2$) as the $\ce{C-O}$ bond cannot break on its own but only breaks as a nucleophile forms a bond to the carbon. It thus follows that the nucleophile attacks from the opposite side. Now, note that the t-butyl group is rather bulky and thus, it would sterically hinder the approach of the nucleophile. However, the ethyl group is less sterically-hindering. Thus, the nucleophile would attack the carbon of the ethyl group, forming iodoethane and tert-butanol.
Without the protonation, the reaction can only proceed in a concerted manner.