Strong Induction Proof: Fibonacci number even if and only if 3 divides index

Solution 1:

Part 1 Case 1 proves $3\mid (k+1)\Rightarrow 2\mid F_{k+1}$, and Case 2 and 3 proves $3\cancel\mid (k+1)\Rightarrow 2\cancel\mid F_{k+1}$. The latter is actually proving the contra-positive of $ 2 \mid F_{k + 1} \Longrightarrow 3 \mid k + 1$ direction.

Part 2 You only need the statement to be true for $n=k$ and $n=k-1$ to prove the case of $n=k+1$, as seen in the 3 cases. Therefore, $n=1$ and $n=2$ cases are enough to prove $n=3$ case, and start the induction process.

Part 3 :)

Part 4 Probably a personal style? I agree having both $n=1$ and $n=2$ as base cases is more appealing to me.