I'm going to introduce some notation to hopefully clear things up.
Let $X_1,X_2,\dots,X_{N_1}$ be the samples from the first Poisson distribution, which are independent, and let $S_{N_1}$ be their sum. Let $Y_1,Y_2,\dots,Y_{N_2}$ be the samples from the second Poisson distribution, which are also independent (and independent of the $X_i$'s), and let $T_{N_2}$ be their sum. The $X_i$'s all have mean and variance $\mu_1$ and the $Y_i$'s all have mean and variance $\mu_2$.
The sample mean is then the random variable:
$$\frac{X_1+\cdots+X_{N_1}+Y_1+\cdots+Y_{N_2}}{N_1+N_2} = \frac{S_{N_1}+T_{N_2}}{N_1+N_2}.$$
You have found the mean of the sample mean (not the sample mean, which is again a random variable) to be
$$E\left[\frac{X_1+\cdots+X_{N_1}+Y_1+\cdots+Y_{N_2}}{N_1+N_2}\right]=\frac{E[S_{N_1}]+E[S_{N_2}]}{N_1+N_2}=\frac{N_1 \mu_1+N_2\mu_2}{N_1+N_2}.$$
To find the standard deviation of the sample mean, first find the variance. The two facts you need to use are:
- $\text{Var}(aX)=a^2\text{Var}(X)$ for any constant $a$ and random variable $X$.
- If $X$ and $Y$ are independent, then $\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y)$.
Thus,
\begin{align*}
\text{Var}\left(\frac{X_1+\cdots+X_{N_1}+Y_1+\cdots+Y_{N_2}}{N_1+N_2}\right) &=\left(\frac{1}{N_1+N_2}\right)^2\text{Var}(X_1+\cdots+X_{N_1}+Y_1+\cdots+Y_{N_2})\\
&=\cdots (\text{use fact 2 now})
\end{align*}
I'll let you finish from here.