The random variables $X,Y,Z$ are mutually independent, and have identical expectation, of $0$, and variance, of $1$. This does not mean they are interchangable; they are still three distinct variables.
For instance, $X$ is independent from $Y$, but clearly not independent from $X$.
Because $\mathsf {Cov}(X,Y)=\mathsf E(XY)-\mathsf E(X)\mathsf E(Y)$ , $\mathsf {Cov}(X,Y)=0$, $\mathsf E(X)=0$, and $\mathsf E(Y)=0$, therefore $\mathsf E(XY)=0$.
Because $\mathsf {Var}(X)=\mathsf E(X^2)-\mathsf E(X)^2$ , $\mathsf {Var}(X)=1$, and $\mathsf E(X)=0$, therefore $\mathsf E(X^2)=1$ .
So obviously $\mathsf E(X^2)\neq \mathsf E(XY)$ and so forth.
So we have : $\mathsf E[(X^2)(Y+5Z)^2]~{=\mathsf E(X^2)~\mathsf E(Y^2+10YZ+25Z^2)\\~\vdots\\ = 26}$
First apply law of total expectation and get
$$\mathbb{E}\left[\frac{X}{N} \right]=\mathbb{E}\left[\mathbb{E}\left[\frac{X}{N}\left|N=n \right. \right] \right]=\mathbb{E}\left[\frac{1}{n}\mathbb{E}[X] \right]=\mathbb{E}\left[\frac{1}{n}\cdot n\theta \right]=\theta$$
then using the definition
$$\mathbb{V}[X]=\mathbb{E}[X^2]-\mathbb{E}^2[X]$$
you can find your variance
$$\mathbb{E}\left[\left(\frac{X}{N} \right)^2 \right]=\mathbb{E}\left[\mathbb{E}\left(\frac{X}{N} \right)^2 |N=n \right]=$$
$$=\mathbb{E}\left[\frac{1}{n^2}\mathbb{E}[X^2] \right]=\mathbb{E}\left[\frac{\theta(1-\theta)}{n}+\theta^2 \right]=\mathbb{E}\left[\frac{1}{N} \right]\theta(1-\theta)+\theta^2$$
which is
$$\mathbb{V}\left[\frac{X}{N} \right]=\mathbb{E}\left[\frac{1}{N} \right]\theta(1-\theta)+\theta^2-\theta^2=\mathbb{E}\left[\frac{1}{N} \right]\theta(1-\theta)$$
...as requested
Best Answer
Note that $$ EY^2=\frac{1}{\sigma^2}E[(X-\mu)^2]=\frac{1}{\sigma^2}\sigma^2=1 $$ by the (original) definition of variance whence $$ \text{Var}(Y)=EY^2-(EY)^2=1-0=1 $$