The problem sounds like this :
How many times should we toss a coin, to say, with a 0.6 probability, that the frequency deviation of tails apparition is lower than 0.01 from 0.5.
My questions are : how is this problem solved ? is there a formula that depends of n , mean , sd and give the wanted probabillities?
In the usual coin flipping model, the number of appearances of tails is binomially distributed, where the parameters are the number of flips $n$ and the probability of a tails appearing $p$. The mean of such a variable is $np$; the variance is $np(1-p)$. The "sample proportion" is this variable divided by $n$, so it has mean $p$ and variance $\frac{p(1-p)}{n}$.
Regardless of any additional assumptions, we can apply Chebyshev's inequality to say that the probability that the sample proportion deviates from $p$ by more than $\delta$ is at most $\frac{p(1-p)}{n\delta^2}$. This is at its largest when $p=1/2$, in which case it is $\frac{1}{4n\delta^2}$. Thus we can ensure that the sample proportion deviates from $p$ by less than $\delta$ with probability at least $1-\alpha$ by taking $n>\frac{1}{4 \alpha \delta^2}$. This is regardless of what $p$ is.
When $n$ is large we can do better. In view of the central limit theorem, if the true proportion is $p$ and $n$ is large enough, the probability that the sample proportion deviates from $p$ by more than $\delta$ is approximately $2\Phi \left ( \frac{-\delta \sqrt{n}}{\sqrt{p(1-p)}} \right )$, where $\Phi$ is the standard normal CDF (which is not an elementary function, but is available in tables and software). This probability is again largest when $p=1/2$, in which case it is $2\Phi(-2\delta \sqrt{n})$. So your sample proportion will be within $\delta$ of $p$ with probability at least $1-\alpha$ if $n$ is at least $\frac{\Phi^{-1}(\alpha/2)^2}{4\delta^2}$ and is large enough that CLT is applicable. ($\Phi^{-1}$ is also available in tables and software.)
Remark: "$n$ is large enough that CLT is applicable" depends rather strongly on what $p$ is. Specifically, the more $p$ deviates from $1/2$, the larger $n$ must be to use CLT. So if $p$ might be very large or small, this calculation might be problematic, in which case it could be desirable to fall back on the first approach.