Let $(N_t)_{t\geq 0}$ be a Poisson process with parameter $\lambda=2.$ Find $\mathbb{E}[N_3N_4].$
The solution here is
\begin{align} \mathbb{E}[N_3N_4]&=\mathbb{E}[N_3(N_4-N_3+N_3)]\tag1\\ &=\mathbb{E}[N_3(N_4-N_3)+N_3^2]\tag2\\ &=\mathbb{E}[N_3(N_4-N_3)]+\mathbb{E}[N_3^2]\tag3\\ &=\mathbb{E}[N_3(N_4-N_3)]+\mathbb{Var}[N_3]+\mathbb{E}[N_3]^2\tag4\\ &=\mathbb{E}[N_3N_1]+\mathbb{Var}[N_3]+\mathbb{E}[N_3]^2\tag5\\ &=\mathbb{E}[N_3]\mathbb{E}[N_1]+\mathbb{Var}[N_3]+\mathbb{E}[N_3]^2\tag6\\ &=6\cdot2+6+6^2=54\tag7 \end{align}
My question is: Why is $N_3$ and $N_1$ independent but not $N_3$ and $N_4$? I assume independence is the reason we can go in (5) from $\mathbb{E}[N_3N_1]$ to (6) where we instead get $\mathbb{E}[N_3]\mathbb{E}[N_1]$?
EDIT/addition:
In the next question in my book, they ask me to explain why the following proof is wrong:
\begin{align} \mathbb{E}[(N_3)^2]=\mathbb{E}[N_3N_3]=\mathbb{E}[N_3(N_6-N_3)]=\mathbb{E}[N_3]\mathbb{E}[N_6-N_3]=\mathbb{E}[N_3]\mathbb{E}[N_3]=\mathbb{E}[N_3]^2. \end{align}
Their answer is that $N_3\neq N_6-N_3$. I find this contradicting because in the previous assignment above, I get to use that $N_4-N_3=N_1$.
$N_3$ and $N_1$ are not independent. (They can't be! Clearly the time until the third point in your Poisson process depends on the time until the first one - for example, trivially $N_3 > N_1$.) But $N_4 - N_3$ doesn't equal $N_1$, either. This is an abuse of notation.
The $N_1$ in line 5 really should be written as something like $\tilde{N}_1$ - it's the first arrival time of the Poisson process starting at time $N_3$. Then you have $E(N_3 (N_4 - N_3)) = E(N_3 \tilde{N}_1) = E(N_3) E(\tilde{N}_1)$. But since $(\tilde{N}_t)_{t \ge 0}$ is again a Poisson process with $\lambda = 2$ you have $E(\tilde{N}_1) = E(N_1)$.