Y is a random variable defined as the sum of N independent Bernoulli trials where the probability of every bernoulli trial equalling '1' is equal to p. The number of Bernoulli trials N is itself a random variable that behaves according to a Poisson distribution function with the parameter lambda.
for part 1) for P(Y | N=n) am I right to assume that I should substitute in the poisson function into the 'N' of the binomially distributed variable Y?
Is this somehow related to the correlation coefficient? Would the correlation coefficient be positive because after all Y is the sum of bernoulli trials thus as N=n increases Y increases?
What about the effect of lambda on E(X)? Should i define lambda as n.p so that as lambda increases, N=n must increase as p is constant?
Hint: $Y$ is the sum of a series of $N$ Bernoulli($p$) random variables; each realising values in $\{0,1\}$, where $N$ is Poisson($\lambda$) Distributed. This then counts the "successes" among $N$ iid Bernoulli trials. Thus $Y$ is Binomially($N, p$) distributed conditional on $N$.
$$Y\mid N ~\sim~ \mathcal{Bin}(N,p) ~\iff ~\mathsf P(Y{=}y\mid N{=}n) ~=~\\ N~\sim~\mathcal{Pois}(\lambda) ~ \iff ~ \mathsf P(N{=}n)~=~$$
Can you now complete and continue?However, you were asked not to derive anything this way.
Instead, please use the Law of Iterated Expectation (sometimes called the Law of Total Expectation). $$\mathsf E(Y) = \mathsf E(\mathsf E(Y\mid N))$$
Use
$$Y\mid N ~\sim~ \mathcal{Bin}(N,p) ~\implies~ \mathsf E(Y\mid N) ~=~ \\ N~\sim~\mathcal{Pois}(\lambda) ~\implies~ \mathsf E(N) ~=~$$
Now, can you now complete and continue?
$\Box$