Consider the process where you spin a quarter a flat surface. If heads
record 1. If tails record 0. 
1. Assuming tht heads and tails are equally likely (null hypothesis)
what is (mu,std. deviation) for the distribution?
2.what is the expected standard deviation (S) for the sample mean if
n=100?
what is the confidence interval level of mean 90,95,99 percent?

The process you describe will generate the well-known binomial
distribution (a Google search on this term will turn up numerous
relevant references).  The binomial distribution governs situations in
which one has a series of trials (i.e., coin tosses) for which there
are only two possible outcomes of each trial (i.e., heads or tails). 
Although in your case, you have specified that the two possibile
outcomes are equally likely, this need not be the case in general.

The general equation for the binomial distribution is:

B(x) = [n! * p^x * (1-p)^(n-x)]/[x! * (n-x)!],

where p is the probability of the first possible outcome, (1-p) is the
probability of the second possible outcome, n is the number of trials,
and x is the number of times the first possible outcome occurs in
those n trials.  B(x) is the probability that the first outcome will
occur exactly x times in the n trials.

If the value of the first outcome is set equal to 1, and the value of
the second outcome set equal to zero, this distribution has a mean
equal to n*p and a standard deviation of sqrt[n*p*(1-p)].  If, as in
your example, p = (1-p) = 0.5, then the mean is simply equal to n/2,
and the standard deviation of the distribution is equal to sqrt[n]/2.

The standard deviation of the mean of a distribution is equal to the
standard deviation of the distribution divided by the square root of
the number of trials.  For the case of a binomial distribution, this
results in: SD-mean = sqrt(n*p*(1-p)/n) = sqrt(p*(1-p)), and is
independent of the number of trials!  For p = 0.5, SD-mean = 0.5.

I am not sure what you mean by "confidence interval level of mean".