How to be able to tell if a molecule will react via SN1 or SN2?
Factors affecting SN1 mechanism:


  
  Weak nucleophile / base and Good leaving group.
  More steric hindrance.
  Polar protic solvent should be used.
  


Factors affecting SN2 mechanism:


  
  Strong nucleophile / base and Good leaving group.
  Less steric hindrance.
  Polar aprotic solvent should be used.
  




Coming back to your problem. Actually, both of the compounds can go through SN1 or SN2 mechanism as the carbon linked with $\ce{Cl}$ is a secondary carbon. But, as you mentioned that you're using $\ce{EtOH}$, and you should know that $\ce{EtOH}$ is little basic, almost neutral, therefore, we can say that now, both of the compounds will go through SN1 mechanism (Remember, the first point of SN1 mechanism, i.e., weak base).

Now, the question is if both the compounds will follow SN1 mechanism, then which will react faster. For this, we will have a look at the SN1 mechanism:




  Step 1: Leaving of the (-X) group
  
  Step 2: Weak nucleophile attacks from either side
  
  Step 3: Substitution product formed (Inversion or retention, based on
  the step 2)


So, our concern is only the step 2.

Note that, rate of reaction in step 2 depends on steric hindrance in your case. So, let's have a look at the structure of (1-chloro-2,2-dimethylpropyl)benzene and (1-chloro-2-methylpropyl)benzene:




The Carbon with red dot is the point where $\ce{EtOH}$ will attack after the (-Cl) will leave.

Now, on comparing the both the compounds, you will notice that in the first compounds there is greater steric hindrance as compared to the second compound. And, that's the reason why second compound will react faster than the first one.

Conclusion:


  (1-chloro-2-methylpropyl)benzene will react faster than
  (1-chloro-2,2-dimethylpropyl)benzene due to less steric hindrance
  around the carbon that is directly attached to aromatic ring (i.e.,
  the carbon with red dot).
  
  And, both of the compounds will go through SN1
  mechanism due to presence of $\ce{EtOH}$.