How do I get the goal $(A \land B) \lor (A \land C)$ from the premises $A \land (B \lor C)$? (Using Fitch) I have 
$$\begin{array}  {r|c:l}
           1. & A \land (B \lor C)
 \\            2. & A \land B
 \\            3. & A          & \land \textsf{ Elim } 2
 \\            4. & B          & \land \textsf{ Elim } 2
 \\            5. & A \land B  & \land \textsf{ Intro } 4, 3
 \\[2ex]\quad 6. & A \land C
 \\            7. & A          & \lor  \textsf{ Elim } 6
 \\            8. & C          & \land \textsf{ Elim } 6
 \\            9. & A \land C  & \land \textsf{ Intro } 8, 7
\end{array}$$
where do I go from here?
How do I get $A \land  B \lor (A \land  C)$
thanks
 A: That is a little muddled, but it kind of looks like you have the right idea.   You've just left the path and wandered into the scrub as soon as you entered the bushwalk.
You need to use $\wedge$ elimination twice on your premise to obtain $A$ and $B\vee C$.   Then you apply case work (first assume $B$, then repeat with $C$), using $\wedge$ introduction (with $A$ and your assumption), $\vee$ introduction to obtain the required conclusion, and to $\to$ introduction to discharge each assumption.   Thus setting things up for $\vee$ elimination.
$$\newcommand{\noshow}[1]{}\begin{align}\because ~& \\ \hdashline & \boxed{\begin{array}{l|l:l}1. & A \wedge (B \vee C) \\ 2. & {A} & \wedge\textsf{ Elim 1} \\ 3. & {B\vee C} & \wedge\textsf{ Elim 1} \\ \hdashline 4.1 & \quad B & \textsf{Assume} \\ 4.2 & \quad \noshow{A\wedge B} & \wedge \textsf{ Intro }2,4.1 \\ 4.3 & \quad \noshow{A\wedge B)\vee (A\wedge C)} & \vee\textsf{ Intro }4.2 \\ \hline 4 & \noshow{B\to (A\wedge B)\vee (A\wedge C)} & \to\textsf{ Intro }4.1,4.3 \\ \hdashline 5.1 & \quad C & \textsf{Assume} \\ 5.2 & \quad \noshow{A\wedge B} & \wedge \textsf{ Intro }2,5.1 \\ 5.3 & \quad \noshow{A\wedge B)\vee (A\wedge C)} & \vee\textsf{ Intro }5.2 \\ \hline 5 & \noshow{C\to (A\wedge B)\vee (A\wedge C)}  & \to\textsf{ Intro }5.3\\ 6 &  (A\wedge B)\vee (A\wedge C)  & \vee\textsf{ Elim } 3,4,5 \end{array}} \\\hline \therefore ~& A \wedge (B \vee C) ~\vdash ~(A\wedge B)\vee (A\wedge C)\end{align}$$
A: Hint 1:  If the only assumption you are given is $A \land B$, can you prove the conclusion?  What about only being given $A \land C$?
Hint 2 : Combine the 2 above proofs with 1 more step from the fitch rules.
Long version:  Looking at the last step, 
$$(A \land B) \lor (A \land C)$$
it seems like the final step might be something like $\lor-\textsf{Intro}$.  But using $\lor-\textsf{Intro}$ would require either establishing  $(A \land B)$ or establishing $(A \land C)$, and neither of those follows from the assumption of $A \lor (B \land C)$.
So looking through the fitch rules, most of them create a theorem with a given structure or would require proving something that would be unprovable.  One candidate stands out as what could be the last step of the proof is $\lor-\textsf{Elim}$:
$$\begin{array} {c}
\begin{array} {c|c|c}
         & x & y \\
x \lor y & \vdots & \vdots \\
         & z & z 
\end{array}
\\ \hline
z
\end{array}$$
because one of the assumptions is a $\lor$ type theorem, and the conclusion does follow from either $B$ or from $C$ (in other words, if we had been asked to prove the conclusion from $A \land X$, it would be true whether $X$ was $B$ or $C$).  So build the proof using $x = B$, $y = C$, $z = (A \land B) \lor (A \land C)$ :
$$\begin{array} {r|l|l}
%
(1) & A \land (B \lor C) & \textsf{Assumption} \\
%
\\
%
(2) & A & \land-\textsf{Elim of } 1 \\
%
(3) & B \lor C & \land-\textsf{Elim of } 1 \\
%
\\
%
(4) & B & \textsf{Assumption} \\
%
(5) & \quad A \land B & \land-\textsf{Intro of } 2 ,~ 4 \\
%
(6) & \quad (A \land B) \lor (A \land C) & \lor-\textsf{Intro of } 5 \\
%
\\
%
(7) & C & \textsf{Assumption} \\
%
(8) & \quad A \land C & \land-\textsf{Intro of } 2 ,~ 7 \\
%
(9) & \quad (A \land B) \lor (A \land C) & \lor-\textsf{Intro of } 8 \\
%
\\
%
(10) & (A \land B) \lor (A \land C) & \lor-\textsf{Elim of } 3 ,~ 4 \to 6 ,~ 7 \to 9
%
\end{array}$$
