jmc

The probability that the second number is more than $\frac{1}{4}$ unit greater than the first number decreases linearly from $\frac{3}{4}$ to $0$ as the first number increases linearly from $0$ to $\frac{3}{4}$. The average of this probability is $\frac{1}{2}\cdot \frac{3}{4}=\frac{3}{8}$. Since there is a $\frac{3}{4}$ chance of choosing a number from $0$ to $\frac{3}{4}$, the probability is $\frac{3}{4}\cdot \frac{3}{8}=\boxed{\frac{9}{32}}$.