Use the Central Limit Theorem to deduce that if $λ$ is large, then $X$ approximately has a normal distribution. 
The time instants of incoming requests at a data server can be modelled with a Poisson process. 
  Let $X$ be the number of requests in one hour and let $λ$ be the intensity (requests per hour) of the Poisson process. 
  - Use the Central Limit Theorem to deduce that if $λ$ is large, then $X$ approximately has a normal distribution. Also specify its parameters.

Can I assume that, if $X\sim\operatorname{Poisson}(\lambda)$ then:
$$
\frac{X-\lambda}{\sqrt\lambda} \overset{\text{distribution}}\longrightarrow N(0,1) \text{ as } \lambda\to\infty \\  $$
It is a correct way to prove it? The parameters for the normal distribution are $\mu=0$ and $\sigma^2=1$ right?

Consider two Poisson iid variables of parameter $\lambda=1$.
Their pdf is $$p_1(k)=\frac1{ek!}.$$
The pdf of the sum of these variables is given by
$$p_2(k)=\sum_{i+j=k}\frac1{e^2i!j!}=\frac1{e^2k!}\sum\binom ki=\frac{2^k}{e^2k!}$$ which is simply a Poisson law of parameter $\lambda=2$.
More generally, you could show that the sum of $\lambda$ Poisson iid variables of parameter $1$ follows a Poisson law of parameter $\lambda$.
Then by the CLT,
$$\frac{p_\lambda-\lambda}{\sqrt\lambda}\to N(0,1)$$
and 
$$p_\lambda\to N(\lambda,\sqrt\lambda).$$
