You cannot, without knowing more about the distribution of $X$.
Suppose for example that $X_1$ has a standard normal distribution, and $X_2$ has a Bernoulli distribution (so that $P(X_2 = 1) = P(X_2 = -1) = 1/2$). Then $X_1, X_2$ both have mean zero and variance 1, but $P(a < X_1 < b)$ is completely different from $P(a < X_2 < b)$. (For instance, if $a=-1$, $b=1$, then $P(a < X_1 < b) \approx 0.683$ but $P(a < X_2 < b) = 0$.)
Intuitively, the mean and variance contain too little information to determine a distribution.