No. For non-negative i.i.d. $Y_i$, $$P(Y_1\ge (\mu+t)n)\le P\Bigg(\sum_{i=1}^n Y_i\ge (\mu+t)n\Bigg)\le\exp(-nf(t))$$ implies that $Y_1$ is sub-exponential and a square of a sub-exponential is not guaranteed to be sub-exponential.
However you can obtain $$P(X_1^2+\dots+X_n^2>nt)\sim nP(X_1^2>nt)=n\exp(-\lambda\sqrt {nt})$$ for $X_i\sim \exp(\lambda)$ which you can extend to all subexponential $X_i$ that have exponential tails.
1) Knowing that in the first four hours $8$ persons got into the bank, calculate the probability of exactly half of them having entered past the first two hours.
I define the random variable $Y_i$ =number of clients in the i-th hour of the opening hours $Y_i$=number of clients in the i-th hour of the opening hours
for $i=1,2,3,4$ , since $Y_i \sim P(λ)$ and these random variables are independent, then $Y=Y_3 +Y_4 \sim P(2λ)$ . The random variable that seems suitable to calculate the probability is $X∼Bin(8,P(Y=4))$.
Fine until the last sentence. Close, but no.
What you have not taken into account is the given condition that $Y_1+Y_2+Y_3+Y_4=8$. You know that eight Poisson point events are distributed across these time intervals. But how are they distributed across these intervals?
Uniformly, that's how. Each of these four events may have occurred in the latter two hours with probability $1/2$, so the count of the events that did so is Binomially distributed, via: $X\sim\mathcal {Bin}(4, 1/2)$
$$\mathsf P(X=4) ~=~ \dbinom{8}{4}\dfrac 1 {2^8} ~=~ \dfrac {35}{128}$$
To verify the argument we work from the premises that (1) the number of clients who arrive in an interval $t$ is Poisson distributed with rate $t\lambda$, and that (2) the count of arrivals in disjoint intervals will be independent.
That is to say: If we let $X_1$ be the count of clients who arrive in the earlier two hours, and $X_2$ be the count of those who do so in the latter two hours. Then $X_1\sim\mathcal{Pois}(2\lambda)$ and independently $X_2\sim\mathcal {Pois}(2\lambda)$. Also $\mathcal X_1+X_2\sim\mathcal{Pois}(4\lambda)$
The conditional probability that $k$ clients arrived in the later two hours when given that $n$ arrived somewhen in the four hours is:
$$\begin{align}\mathsf P(X_2=k\mid X_1+X_2=n) ~=~ &\dfrac{\mathsf P(X_2=k)~\mathsf P(X_1=n-k)}{\mathsf P(X_1+X_2=k)}
\\[2ex] ~=~ & \dfrac{\dfrac{(2\lambda)^k~\mathsf e^{-2\lambda}}{k!}~\dfrac{(2\lambda)^{n-k}\mathsf e^{-2\lambda}}{(n-k)!}}{\dfrac{(4\lambda)^n\mathsf e^{-4\lambda}}{n!}}
\\[2ex] ~=~ & \dfrac{n!}{k!~(n-k)!}\dfrac 1{2^n}
\end{align}$$
Best Answer
Hint
I don't think you need to do any summing. You have $$ P(X\ge t) > \frac{e^{-\lambda}\lambda^t}{t!}.$$ You can apply Stirling to this to get an asymptotic lower bound and show it decays slower than any Gaussian.