Solution:

Since $p_{4} - p_{1} = 8$, and no prime is even, we observe that $\{p_{1}, p_{2}, p_{3}, p_{4}\}$ is a subset of $\{p_{1}, p_{1} + 2, p_{1} + 4, p_{1} + 6, p_{1} + 8\}$. Moreover $p_{1}$ is larger than $3$. If $p_{1} \equiv 1 \pmod{3}$, then $p_{1} + 2$ and $p_{1} + 8$ are divisible by $3$. Hence we do not get $4$ primes in the set $\{p_{1}, p_{1} + 2, p_{1} + 4, p_{1} + 6, p_{1} + 8\}$. Thus $p_{1} \equiv 2 \pmod{3}$ and $p_{1} + 4$ is not a prime. We get $p_{2} = p_{1} + 2$, $p_{3} = p_{1} + 6$, $p_{4} = p_{1} + 8$.

Consider the remainders of $p_{1}, p_{1} + 2, p_{1} + 6, p_{1} + 8$ when divided by $5$. If $p_{1} \equiv 2 \pmod{5}$, then $p_{1} + 8$ is divisible by $5$ and hence is not a prime. If $p_{1} \equiv 3 \pmod{5}$, then $p_{1} + 2$ is divisible by $5$. If $p_{1} \equiv 4 \pmod{5}$, then $p_{1} + 6$ is divisible by $5$. Hence the only possibility is $p_{1} \equiv 1 \pmod{5}$.

Thus we see that $p_{1} \equiv 1 \pmod{2}$, $p_{1} \equiv 2 \pmod{3}$ and $p_{1} \equiv 1 \pmod{5}$. We conclude that $p_{1} \equiv 11 \pmod{30}$.

Similarly $q_{1} \equiv 11 \pmod{30}$. It follows that $30$ divides $p_{1} - q_{1}$.