$N(t)$ is a Poisson process with rate $\lambda$ and is independent of the sequence $X_1,X_2,...$ of independent and identically distributed random variables with mean $\mu$ and variance $\sigma^2$.
Find Cov$(N(t),\sum_{i=1}^{N(t)}X_i)$. I have a general idea of how to find this, but need some help filling in the gaps.
$$Cov(N(t),\sum_{i=1}^{N(t)}X_i)=E[N(t)\sum_{i=1}^{N(t)}X_i]-E[N(t)]E[\sum_{i=1}^{N(t)}X_i]$$
I know that $E[N(t)]=\lambda t$.
I know that $E[N(t)\sum_{i=1}^{N(t)}X_i]=E[E[N(t)\sum_{i=1}^{N(t)}X_i|N(t)]]=E[(N(t))^2\mu]=\mu E[(N(t))^2]$
I think (but am not sure about this) that $E[\sum_{i=1}^{N(t)}X_i]=E[X_1]+E[X_2]+\cdots+E[N(t)]=\mu\cdot N(t)$
I am not sure how to find $E[(N(t))^2]$, and then also would like to know if my previous statements are correct or not.
Let $S(t)=\sum_{i=1}^{N(t)}X_i$ . The expectation of $S(t)$ is easy to calculate by $E[S(t)] = E\left( E[S(t)\mid N(t)]\right)$ which is $\mu E[N(t)] = \mu\lambda$, by the same reasoning as your last "know that" formula.
As for $EN(t)^2$. The rv $N(t)$ is Poisson with parameter $t\lambda$, so has expectation $t\lambda$ and variance $t\lambda$ and hence second moment $t\lambda +(t\lambda)^2$.