Is there only one right solution when solving with Fitch system? I was wondering if there was only one dead-set answer for each Fitch proofs. Are solutions that are given by the textbook the only way to answer a Fitch proof question (do all the steps have to look exactly the same?) ?
 A: No. If there exists one Fitch proof, there always exist infinitely many. 
One reason is that it is always possible to add useless steps: For example, given the derivation 

we could also have done this:

In the second proof, we added a $\land$ introduction directly followed by a $\land$ elimination. Steps of an introduction rule immediately followed by an elimination rule for the same connective are always redundant and can be removed (this process is called normalization). But still, the second proof is just as well a valid Fitch proof of $P \to Q, P \vdash Q$. And it is obvious that we can add usesless steps however many times we want, and thereby get infinitely many possible Fitch proofs for the sequent $P \to Q, P \vdash Q$.  
But we don't have to add redundancy. For example, consider the following derivation of $P \to Q \lor Q, P \vdash Q$:  
 
Note that the $\lor E$ comes before the $\to I$ step.  
But we can also do this:  

which is a valid proof of the same sequent but with the order of the $\to I$ and $\lor E$ applications swapped around.  
As you can see: There often is more than one reasonable way to construct a proof, and with redundant steps we can in fact create infinitely many of them.
