Let $p$ be a cute prime. The numbers $pq - 2$, $pq$ and $pq + 2$ give all the possible remainders modulo $3$, so one of them must be divisible by $3$. If $3 \mid pq - 2$, then $pq - 2 = 3$ and $pq = 5$, which is impossible, as $p$, $q$ are primes. Similarly if $3 \mid pq + 2$, then $pq + 2 = 3$ and $pq = 1$, which is also impossible. So $3 \mid pq$, which means that either $p = 3$ or $q = 3$. Given that in the case $p = 3$ we can have $q = 3$, we have shown that $p$ is cute iff both $3p - 2$ and $3p + 2$ are primes.

Therefore $p$ is wonderful if $p$, $p + 2$, $3p - 2$, $3p + 2$, $3p + 4$ and $3p + 8$ are all primes. We know that the numbers $3p - 2$, $3p$, $3p + 2$, $3p + 4$ and $3p + 6$ give all of the possible remainders modulo $5$, so one of those numbers is divisible by $5$. If it's $3p - 2$, $3p + 2$ or $3p + 4$, then due to primality it must be equal to $5$, but then $p$ is not prime. If it's $3p$ or $3p + 6$, then either $5 \mid p$ or $5 \mid p + 2$ respectively. Thus $p = 3$ or $p = 5$. We can verify that both work.