Sure, here is the translated text:

```
II. Try to find all prime numbers $p$ that satisfy the following conditions:
For any prime number $q<p$, if $p=b_{q}+r, 0 \leqslant r<\dot{q}$, then there does not exist an integer $a$ greater than 1 such that $a^{i}$ divides $r$.
```
Solution: It is easy to verify that $p=2,3,5,7$ all satisfy the conditions. Now let's discuss the prime $p \geqslant 11$. If $p$ satisfies the conditions, then we have:
(i) $p-4$ has no prime factors greater than 4;
(ii) $p-8$ has no prime factors greater than 8;
(iii) $p-9$ has no prime factors greater than 9.
From (i) and the fact that $p-4$ is odd, we get
$$
p-4=3^{a}, a \geqslant 2 \text{. }
$$

From the above equation and the fact that $p-8$ is odd, we know that $p-8$ cannot be divisible by 2 or 3. Therefore, from (ii) we have
$$
p-8=5^{b} 7^{c} \text{. }
$$

From equations (1) and (2), we get
$$
5^{b} 7^{c}-3^{a}+4=0, a \geqslant 2 \text{. }
$$

Dividing both sides of equation (3) by 3, we get $(-1)^{b}+1=0$, so,
$1 \leqslant b=2 l+1, l \geqslant 0$.
Since $b \geqslant 1$, from equation (3) we deduce that 5 divides $3^{a}+1$, so,
$2 \leqslant a=4 m+2, m \geqslant 0$.
From equation (2), we know $p-9=5^{b} 7^{c}-1$. It is clear that $p-9$ cannot be divisible by 3; since $b \geqslant 1$, it also cannot be divisible by 5.
We will prove that $c=0$ must hold.
If not, suppose $c>0$, then $p-9$ also cannot be divisible by 7. Therefore, from (iii) we must have
$$
p-9=5^{b} 7^{\dot{c}}-1=2^{d}, c>0 \text{. }
$$

This implies that 7 divides $2^{d}+1$, which is impossible for any non-negative integer $d$ (why?).
From equations (1), (2), and $c=0$, we get
$$
5^{b}=3^{a}-4=\left(3^{2 m+1}-2\right)\left(3^{2 m+1}+2\right) \text{. }
$$

Let the greatest common divisor of $3^{2 m+1}-2$ and $3^{2 m+1}+2$ be $g$. Clearly, $g$ is odd, and $g$ divides $\left(3^{2 m+1}+2\right)-\left(3^{2 m+1}-2\right)=$
4, so $g=1$. This, along with the fact that 5 is a prime, implies
$$
3^{2 m+1}-2=1,3^{2 m+1}+2=5^{b} \text{. }
$$

Thus, $m=0, b=1, a=2$.
Therefore, from equation (1), the only prime that satisfies the conditions is 13.
The complete set of primes that satisfy the conditions of this problem is: $2,3,5,7,13$.