Solution:

a) Set $S_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$. We have that $S_{2}=\frac{3}{2}$, $S_{7}=\frac{3121}{140}$,
$$\begin{aligned} S_{22}-S_{7}= & \frac{1}{8}+\frac{1}{22}+\frac{1}{10}+\frac{1}{20}+\frac{1}{11}+\frac{1}{19}+\frac{1}{14}+\frac{1}{16}+\frac{1}{13}+\frac{1}{17}+ \\ & \frac{1}{9}+\frac{1}{12}+\frac{1}{15}+\frac{1}{18}+\frac{1}{21}=\frac{30 a}{b}+\frac{51}{140}=\frac{3 c}{d} \end{aligned}$$
where $(a, b)=(c, d)=1$. It is easy to see that $a \equiv b \pmod{3}$. Hence $3 \nmid c, d$, $c \not \equiv d \pmod{3}$, $p_{22}=3 p_{22}'$ and $3 \nmid p_{22}'$. Similarly, we have that $S_{67}-S_{22}= \frac{90 e}{f}+\frac{c}{d}$, where $3 \nmid f$. It follows that $3 \nmid p_{67}, q_{67}$.

b) Set $S_{n}=\frac{k_{n}}{3^{m_{n}} l_{n}}$, where $3 \nmid k_{n}, l_{n}$. Then
$$\begin{aligned} S_{3 n} & =\frac{S_{n}}{3}+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\cdots+\frac{1}{3 n-2}+\frac{1}{3 n-1} \\ & =\frac{k_{n}}{3^{m_{n}+1} l_{n}}+3 \cdot \frac{a_{n}}{b_{n}}=\frac{k_{n} b_{n}+3^{m_{n}+2} l_{n} a_{n}}{3^{m_{n}+1} l_{n} b_{n}} \end{aligned}$$
where $3 \nmid b_{n}$. Therefore, if $m_{n} \geq -1$, then $m_{3 n}=m_{n}+1$. Analogously, we have that $m_{3 n+2}=m_{n}+1$ for $m_{n} \geq -1$ and $m_{3 n+1}=m_{n}+1$ for $m_{n} \geq 0$. Since $m_{1}=0$, $m_{2}=m_{7}=m_{22}=-1$ and $m_{67}=0$, it is easy to see that the answer is $n=2,7,22$.