We want A1⇒A2⇒A95 A3 to yield exactly:

1 1  
37 30 14 31 36 32 35 34 33 1 15 28 29 30  
27 18 17 21 22 23 24 25 26 1 16 20 19 18  
13 3 12 5 9 6 10 8 7 1 11 4 2 3

The first two 1’s come from A95→A97 A96.  
The middle segments come from the A3→A27→A28→A36→A32 A37 path, with A32 producing the “37 30 … 30” block and A37 producing the “27 18 … 18” block, and finally A43 producing the last “13 3 12 5 … 2 3” block.  

Comparing to the existing productions we must insert missing terminals:

1) In rule_1036 (A32→37 30 31 A25 15 28 29) we need a 14 before the 31, and a trailing 30:  
   A32→37 30 14 31 A25 15 28 29 30  

2) In rule_1059 (A37→27 18 17 21 A23 20 18) we must insert 16 before the 20 and 19 before the final 18:  
   A37→27 18 17 21 A23 16 20 19 18  

3) In rule_1090 (A43→13 3 12 A21 11 2) we must insert 5 before A21, 4 before the 2, and a final 3:  
   A43→13 3 12 5 A21 11 4 2 3

These three changes (7 new terminals total) allow the grammar to generate exactly the target sequence.  

Updated production rules:  
rule_1036: A32 -> 37 30 14 31 A25 15 28 29 30  
rule_1059: A37 -> 27 18 17 21 A23 16 20 19 18  
rule_1090: A43 -> 13 3 12 5 A21 11 4 2 3