To show that they are unbiased, you need to show that the expectation of the sample mean is equal to the parameter.
In a Bernoulli distribution with n samples, you know that the mean of a Bernoulli is simply Successes/Outcomes = p where p is a constant in this particular example.
Now the sample mean is given by [X1 + X2 + ... + XN]/N but all the X's are independent random Bernoulli random variables.
To be unbiased, you need to show that E[x_bar] = p. The first thing to figure out is what the expectation of a single observation is and use that to prove the result.
The variance is pretty much the same, but for this you want to use the fact that the sum of independent Bernoulli's with the same parameter is Binomial distribution with parameters n and p. So you know that [X1 + X2 + ... + XN]/N = 1/N * Binomial(n,p) and Var[1/N * X] = 1/N^2 * Var[X].