To solve this problem, we need to analyze the given statistics about the observations \(x_1, x_2, \ldots, x_{20}\) in the interval \([0, 1]\) and evaluate each given statement.

Statement 1: If 15 observations are smaller than 0.3, then \(\bar{x}\) cannot exceed 0.5.

Since 15 observations are below 0.3, these will contribute a maximum to the mean as \(15 \times 0.3 = 4.5\). Let the remaining 5 observations be at most 1 (maximum value in the interval). So, their contribution is \(5 \times 1 = 5\).

Thus, the total sum \(S = 4.5 + 5 = 9.5\). Therefore, the mean is:

\(\bar{x} = \frac{S}{20} = \frac{9.5}{20} = 0.475\)

Since \(0.475 \leq 0.5\), this statement is true.

Statement 2: \(s^2\) will be maximum if 10 of these observations are 1 and the rest are 0.

If 10 observations are 1, and the rest 10 are 0, then we have a setup like:

\(x_1 = x_2 = \cdots = x_{10} = 1, \quad x_{11} = x_{12} = \cdots = x_{20} = 0\)

Here, \(\bar{x} = \frac{10 \times 1 + 10 \times 0}{20} = 0.5\).

Variance is calculated as:

\(s^2 = \frac{1}{20} \left[10 \times (1 - 0.5)^2 + 10 \times (0 - 0.5)^2\right] = \frac{1}{20} (10 \times 0.25 + 10 \times 0.25) = 0.25\)

This variance is indeed the maximum possible variance for binary observations in [0, 1], validating the statement.

Statement 3: If all observations except one are smaller than 0.5, then \(\bar{x}\) cannot be smaller than \(\tilde{x}\).

This statement is not necessarily true. Consider 19 observations close to 0 and one being close to 1. The mean can be less than the median if the minority element is heavily skewed towards 1, thereby invalidating this claim.

Statement 4: \(s^2 \leq \bar{x}(1 - \bar{x})\).

This is a true statement and follows from the properties of variance for any distribution in the interval \([0, 1]\). The binomial distribution leads to maximum variance of \(pq\) when the mean is \(p \leq 1\), fitting perfectly for our conditions.

Therefore, the correct answers are Statement 1, Statement 2, and Statement 4.