Reductio ad Absurdum within a Conditional Proof I am trying to formulate a proof for:
1. ~Q → (L → F)  
2.  Q → ~A  
3.  F → B  
4.  L           ∴ ~A v B
---------------------------
5.  ~Q       Assume for(CP)
6.   L → F   1,5. MP
7.   L → B   3,4,6. HS
8.   B       4,7. MP
9.  ~A • B   Assume(For RAA)
10. ~A       Simp
11.  A       2,5. MT
12.  A • ~A  Add
13. ~A v B   9-12 RAA, 5-13 CP

Would you put RAA and CP in the same line like this?
 A: 

  
*is supposed to be Q -> ~A
  

In that case, one approach might be to just go for a reductio ad absurbum in the following way:
    1. ~Q → (L → F)  
    2.  Q → ~A  
    3.  F → B  
    4.  L           ∴ ~A v B
    ---------------------------
    5.  ~(~AvB)  Assume for RAA
    6.      A • ~B  5 DeM
    7.      A       6 simp
    8.      ~B      6 simp
    9.      ~F      3,8 MT
    10.     ~Q      2,7 MT
    11.     L → F  1,10 MP
    12.     F       4,11 MP
    13.     ~F • F  9,12 Add 
    14. ~A v B   5-12 RAA

Another pretty fun approach would be to do a conditional proof of $A\to B$ and then use a conditional exchange to obtain $\lnot A \lor B$. It would be a good exercise to try that out yourself
A: With help from the comments:
1. ~Q → (L → F)  
2.  Q → ~A  
3.  F → B  
4.  L           ∴ ~A v B
---------------------------
5.  ~Q       Assume for(CP)
6.   L → F   1,5. MP
7.   L → B   3,4,6. HS
8.   B       4,7. MP 1-8 CP
9.  ~A v B   Add

A: 
i messed up 2. is supposed to be Q -> ~A

Then there you do not require RAA.  Use a proof by cases.
$\begin{array}{r:ll}1 & \lnot Q \to (L \to F)\\  
2&  Q \to \neg A \\ 
3 &  F \to B \\  
4 &  L  \\\hdashline
5 & \quad \neg Q &\text{Assume} \\
6 & \quad L\to F & 1,5, \to\text{Elim, aka modus ponens} \\
7 & \quad F & 4,6 , \to\text{Elim}\\
8 & \quad B & 4,7 , \to\text{Elim}\\
9 & \quad \neg A\vee B & 8, \vee \text{Intro, aka Addition}\\\hline
10 & \neg Q\to (\neg A\vee B) & 5,9, \to\text{Intro, aka Deduction Theorem}\\ \hdashline
11 & \quad Q & \text{Assume}\\
12 & \quad \neg A & 2, 11, \to\text{Elim}\\
13 & \quad \neg A\vee B & 12, \vee\text{Intro}\\ \hline
14 & Q\to (\neg A\vee B) & 11,13,\to\text{Intro}\\
15 & Q\vee\neg Q & \text{Law of Excluded Middle}\\
16 & \neg A\vee B & 10,14,15 \vee\text{Elim, aka proof by cases}
\end{array}$
PS: Constructive Dilemna would have also sufficed.   $5,8$ deduces $\neg Q\to B$ and with $2$ we have $\{Q\to\neg A, \neg Q\to B\}\vdash \neg A\vee B$

Update: Of course, you may use RAA.  Assume the negation and derive a contradiction.   This may be nested: further assume Q, derive a contradiction, thus deducing $\neg Q$ from which a contradiction may also be derived...
$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{~~1.~\neg Q \to (L \to F)\\~~2.~Q \to \neg A\\~~3.~F \to B\\~~4.~L}{\fitch{~~5.~\neg(\neg A\vee B)}{\fitch{~~6.~Q}{~~7.~\neg A\hspace{10ex}\to\mathsf E, 6,2\\~~8.~\neg A\vee B\hspace{6ex}\vee\mathsf I, 7\\~~9.~\bot\hspace{12ex}\neg\,\mathsf E,5,8}\\10.~\neg Q\hspace{14ex}\neg\,\mathsf I, 6{-}9\\11.~L\to F\hspace{10ex}\to\mathsf E, 10,1\\12.~F\hspace{15ex}\to\mathsf E, 4,11\\13.~B\hspace{15ex}\to\mathsf E, 3, 12\\14.~\neg A\vee B\hspace{10ex}\vee\mathsf I,13\\15.~\bot\hspace{16ex}\neg\,\mathsf E, 5,14}\\16.\neg\neg(\neg A\vee B)\hspace{10ex}\neg\,\mathsf I, 5{-}15\\17.~\neg A\vee B\hspace{14ex}\neg\neg\,\mathsf E,16}$
