If we allow exor gates then here is a solution.
1st col 4th map: (A exor C) + AB
2nd col 2nd map: (A exor B) + ABC
2nd col 3rd map: !(A exor C) + AB, alternatively (A exnor C) + AB
Edit:
Take col 1 map 4 for example. The first two columns have a pattern I recognize as an XOR gate. First row is 01, second row is 10. Now I look at the boxes with '1', A changes when jumping between the two. Look for another variable that is also changing; in this case it is C. A and C are either 01 or 10 in the '1' boxes: that is the characteristics of an XOR gate. Only two '1' remain in the map and they are grouped into the term AB.
Now col 2 map 2: Here are two groups that will work for an XOR gate; col 1 & 4, and col 3 & 4. The first group produces an XOR gate, the second an XNOR gate. In the XNOR gate, both inputs must be equal to produce a '1' on the output.
Finally col 2 map 3. This map looks like col 1 map 4, except the XOR pattern is reversed. That means we use an XNOR gate instead of the XOR in the first map, or we add an inverter to the XOR output.
By the way, regarding your equation for col 1 map 4, notice that the two bottom corners both have '1'. You can group them to produce the term A!C, reducing your third term to two variables.
Best Answer
The K-Map is producing either the minimal Sum Of Products (SOP), if "circling" the ones, or the minimal Product of Sums (POS), if circling the zeros. Let's do the latter (my grouping is using the red line):
Now, the group of 4 is giving us \$I\$ (as it is the only term which has the same value of \$0\$ in this group) and the group of two is giving us \$A+B\$. So the final result is \$I\cdot(A+B)\$.