Show that if A is regular so is $A^{R}$
Why do I have to make it as an NFA instead of making an DFA...


You don't have to, but making an NFA allows you to use $\epsilon$ transitions, which are convenient for making your procedure.


Can i kill the state q3...


Sure, but it wouldn't be useful for your proof.  Dead states don't hurt anything, and it would be harder to write this step into your procedure.


Why the first state of the NFA transitions with ϵ?


Because the FMS you are working from may have had multiple final states, so you need a way to encode multiple potential starting points.  Using $\epsilon$ transitions gives you a choice of multiple states.

What you're trying to do is make a procedural proof.  In essence, you're trying to show that for every FSM $A$, you can follow some formula to construct a FSM that also accepts $A^R$.  If you can always construct a Finite State Machine for $A^R$, then $A^R$ must be Regular.

The machine that results from your process doesn't have to be the most efficient machine or contain the minimum possible number of states.  This isn't a work of art, it's a proof of concept: if you can always make an appropriate reversal machine, then your reversal must always be regular. All it must do is work every time to accomplish your proof.