5. For a real number $a>1$, written as a reduced fraction $a=\frac{q}{p}$, $(p, q)=1$. The number of values of $a$ that satisfy $p q=30!$ is $\qquad$.


5.512.
30! has 10 prime factors, $\left(30!=2^{26} \times 3^{14} \times 5^{7} \times 7^{7} \times 11^{2} \times 13^{2} \times 17 \times 19 \times 23 \times 29\right),(p, q)=$ 1. Each prime factor $p_{i}^{\circ}$ is either all in the numerator or all in the denominator, which gives $2^{10}$ cases in total, with those greater than 1 accounting for exactly half. Therefore, the number of $a$'s where $a > 1$ is $2^{9}=512$.