Notice that the max sum of a pair is $19$ while the min sum is $3$. Listing out the prime numbers between $3$ and $19$, we have $3,5,7,11,13,17,19$. Then $3+5+7+11+13+17+19=75$. If we sum all of the pairs, we have $1+2+3+4+5+6+7+8+9+10=55$. Since all the pairs have distinct prime sums, there are two prime numbers in the list that are not used, and sum to $20$. The possible pairs are $(3,17)$ and $(7,13)$. $17$ and $19$ cannot both be sums, since an $8$ must be present in both of them. Thus, the two prime numbers not used are $3$ and $17$. This yields the solution set $\{(10,9),(7,6),(8,3),(5,2),(4,1)\}$.

The max sum of a pair is $39$ while the min sum is $3$. Finding all of the prime numbers between $3$ and $39$, we have $3,5,7,11,13,17,19,23,29,31,37$ Which consists of $11$ prime numbers. Thus, one prime will not be used. Notice that the sum of the $11$ primes is equal to $195$, while the sum of all the sums is $1+2+3+4+...+17+18+19+20=210$. $210>195$, so this arrangement is not possible.