What does it mean if I am unable to write RS configuration for a compound?
I think I get what your question is and will try explain the things to you.


  I was unable to write RS configurations for 1 and 3, since they have carbon atoms with more than one same groups attached to them (they are achiral). Does that mean they are optically inactive?


The crux of your question lies apparently in two problems:


you assume that molecules 1 and 3 are optically inactive, but are unable to visualize this with a center/plane of symmetry
you seem to assume that molecules without chiral carbons are optically inactive


So, let's tackle this:


  Correct me if I am wrong, these compounds don’t seem to have planes or centres of symmetry. 


Well, you're wrong. Note that in simple saturated hydrocarbons (with or without halogen substituents), it is usually safe to assume that if no chiral carbons are present, then the molecule will be optically inactive (without the need to validate this assumption by manually drawing out their planes/centers of symmetry). Here are two examples:


 (source) 

notice how the plane of symmetry cuts-through through the blue and green atoms, using the remaining two grey identical atoms to obtain a plane of symmetry, to prove the molecule is optically inactive. This was expected, since the molecule had no chiral carbon.


it is possible to visualize the plane of symmetry in 2-chloro-2-methylbutane from its eclipsed Newmann projection, as seen above. The plane of symmetry cuts-through the chlorine atom and the methyl radical in front of it, rendering the molecule optically inactive, as expected.



  Are all molecules without any chiral carbon optically inactive?


No, this is wrong. Common examples like ortho-disubstituted-biphenyls (tl;dr: they are propeller shaped rings), allenes (tl;dr: their terminal groups are in perpendicular planes), etc. have no chiral carbons, yet are optically active. I won't detail these as you're still a "beginner" in this, but you'll definitely come across these examples in the very end of your course on optical activity.



In regards to your edited question, it is unreasonable to draw Fischer projection for achiral carbon atoms. For example, when drawing the Fischer projection of carbohydates, the terminal achiral carbon is written as a $\ce{-CH2OH}$ substituent to the penultimate chiral carbon, and not expanded out into its own Fisher projection.

I hope it helps!