Calculate $P(210 \le G_{100} \le 230)$ with the help of the central limit theorem 
Assume you have a fair coin. Everytime you flip tail, you receive $4$€, >otherwise you receive $0$€. For $n \in \Bbb N, n > 0$, let $G_n$ be the profit >when you flipped the coin $n$ times.
Determine the expected value and the variance of $G_n$.
Then, let $n = 100$, and calculate
$$P(210 \le G_{100} \le 230)$$
with the help of the central limit theorem.
Hint:
$$\phi\left({-1 \over 2}\right) = 0,3085, \ \phi\left({-3 \over 2}\right) = 0,0668.$$

We have that
$$G_n = 4S_n$$ with
$$S_n := X_1 \ + \ ... \ +  \ X_n$$
and
$$X_k := 1,$$ if the $k$-th flip shows tail, and $0$ otherwise.
I believe that $G_n$ is binomial distributed. Hence, the expected value and the variane can be calculated like this:
$$E(G_n) = E(4S_n) = 4E(S_n) = 2n,$$
$$Var(G_n) = Var(4S_n) = 16Var(S_n) = 4n \rightarrow \sigma = 2\sqrt n.$$
But how do I have to apply the central limit theorem now? My calculations didn't bring me near the values that would allow me to use the hints.
 A: You went too far by analyzing the distribution of $G_{100}$ ! The goal of the central limit theorem is to avoid this computation and only rely on the expected value and variance of a single $X_i$ (assuming $i.i.d$, independence...).
In your case we can define $Y_i$ being the gain at round $i$ and we compute :
$$
E[Y_i] = 2, \ \ \ \ \ Var(Y_i) = 16\cdot \frac12 -4 = 4 
$$
Therefore the CLT tells you that (for large $n$)
$$G_n = \sum_{i=1}^n Y_i \sim N(2n, 4n)$$
Note that we make no assumption on how $G_n$ is distributed. Sure in your case it is actually binomial since $Y_i$ are ($4 \cdot$) Bernoulli but it doesn't matter. Assuming that the $Y_i$ had a very weird distribution, it could be really hard to compute the exact distribution of their sum.

Extending a bit my answer : You only have at you disposal $\Phi$ which is all about $N(0,1)$ however $G_n$ is $N(2n,4n)$ so you need to recenter it. This is done by defining :
$$
H_N := \frac{G_n -2n}{\sqrt{4n}} \sim N(0,1)
$$
Now this transformation transforms the probability you're looking for :
$$
P(210 \leq G_n \leq 230) = P(\frac{210- 2n}{2\sqrt{n}} \leq H_n \leq \frac{230- 2n}{2\sqrt{n}})
$$
Plugging the value for $n=100$ gives :
$$
P(210 \leq G_{100} \leq 230) = P(\frac12 \leq H_{100} \leq \frac32)
$$
Now you can make use of $\Phi$ since $H_{100} \sim N(0,1)$ !

Expanding more my answer based on the comment of the OP :
In my example I only computed the expected value and variance for a single random variable $Y_i$, not for the whole sum. Let's say I computed $E[Y_i] = \mu, \ Var(Y_i) = \sigma^2$ for a single $Y_i$,
then we know that by linearity of expectation and some basic properties of the variance (Assuming i.i.d) which are absolutely not related to CLT that 
$$
E[\sum_{i=1}^n Y_i] = \sum_{i=1}^n E[Y_i] = n\mu\\
Var(\sum_{i=1}^n Y_i) = \sum_{i=1}^n Var(Y_i) = n\sigma^2
$$
(I let you Google/Wikipedia these properties)
Therefore you know that $G_n$ must have expectation $n\mu$ and variance $n\sigma^2$ and, again, this has nothing to do with de CLT. However you have no idea on the actual distribution of $G_n$, you only know its expectation and variance. What the CLT actually tells you is that $G_n$ is (for large $n$) normally distributed. And since we know that $E[G_n] = n\mu$ and $Var(G_n) = n\sigma^2$ then we can conclude that it must be $\sim N(n\mu, n\sigma^2)$.
A: Following your notations, we have
$$
\Pr(210 \leq G_{100} \leq 230) = \Pr(\frac{105}{2} \leq S_{100} \leq \frac{115}{2})
$$
By central limit theorem, for sufficiently large $n$, $\frac{S_n - n / 2}{\sqrt{n}/{2}}$ can be approximated by $\mathcal{N}(0, 1)$. Therefore,
$$
\Pr(\frac{105}{2} \leq S_{100} \leq \frac{115}{2}) = \Pr(\frac{1}{2} \leq \frac{S_{100} - 50}{5} \leq \frac{3}{2}) = \Phi(\frac{3}{2}) - \Phi(\frac{1}{2})
$$
A: Hint: 
Now you can use that the Standard Normal distribution is symmetric around the mean $0$.

$1-\Phi(z)=\Phi(-z)$
with $z=\frac{x-2n}{2\sqrt n}$ and $n=100$

