The existing answer seems to meet all the requirements in your wish list. For completeness, I just thought I'd add an extremely simple case matching the main requirement, without using continuity, etc. This example does not meet your criteria, but it is added to show that it is extremely easy to get distributions that meet the main requirement (which may be counter-intuitive to some readers).

Example: If $X \sim \text{Bern}(\theta)$ with non-degenerate parameter $0<\theta<1$ you get:

$$\mathbb{P}(X > \mathbb{E}(X)) = \mathbb{P}(X > \theta) = \mathbb{P}(X = 1) = \theta.$$

Taking $\theta = 0.74$ then gives you the required outcome $\mathbb{P}(X > \mathbb{E}(X)) = 0.74$.

The existing answer seems to meet all the requirements in your wish list. For completeness, I just thought I'd add an extremely simple case matching the main requirement, without using continuity, etc. This example does not meet your criteria, but it is added to show that it is extremely easy to get distributions that meet the main requirement (which may be counter-intuitive to some readers).

Example: If $X \sim \text{Bern}(\theta)$ with non-degenerate parameter $0<\theta<1$ you get:

$$\mathbb{P}(X > \mathbb{E}(X)) = \mathbb{P}(X > \theta) = \mathbb{P}(X = 1) = \theta.$$

Taking $\theta = 0.74$ then gives you the required outcome $\mathbb{P}(X > \mathbb{E}(X)) = 0.74$.