Karnaugh map: mirroring tables

Assuming that \$\Phi\$ is the same as X (don't care), I would make the following corrections:

JA = ¬B ¬C ¬D
Here you covered a 0 that you shouldn't, so you have to make a smaller minterm.

KA = D
Here you can make a bigger minterm because the 2*2 are above the minterm only contains don't cares.

KC = ¬D
Again, you can meg a bigger minterm. You don't have to add C because the first column only contains don't cares and ¬D covers the first and the last column.

Lets say your five variables are A, B, C, D, E. To "draw" a 5 variable K-map, draw two K-Maps and imagine them stacked on top of one-another.

E=0

                !C          C
            ----------  ---------
     AB\CD  00    01    11   10
!A { 00     0000  0001  0011 0010  > !B
!A { 01     0100  0101  0111 0110  } B
A  < 11     1100  1101  1111 1110  } B
A  < 10     1000  1001  1011 1010  > !B
            vvvv  ---------- vvvv         
             !D       D       !D      


E=1
                !C          C
            ----------  ---------
     AB\CD  00    01    11   10
!A { 00     0000  0001  0011 0010  > !B
!A { 01     0100  0101  0111 0110  } B
A  < 11     1100  1101  1111 1110  } B
A  < 10     1000  1001  1011 1010  > !B
            vvvv  ---------- vvvv         
             !D       D       !D      

You see how every grid location (which maps to a min-term) differs from all its neighbors (including wrap-around edge-cases) by exactly one bit? That's more-or-less the whole point of a K-Map representation. Now in a 4-variable K-Map you can "circle" contiguous rectilinear shapes where the output is a "1" so long as their "area" is a power of two (i.e. 1, 2, 4, 8, or 16 terms) in order to achieve a "cover" of the K-map. I believe there are 2^16 possible covers of a 4-variable K-map (i.e. 2^(2^4)).

The only difference in the 5-variable case is "adjacent" also include analogous positions in the two 4-variable maps (sub-spaces), and there are more ways to ways to "circle" a number of cells that is a power of two; that is to say there are "three-dimensional" covers possible.

Hopefully it's clear that "the thing that is constant in a straight line" is what that line identifies, and that the intersections of "the things that are constant" is what makes a particular location a minterm. If you "circle" the upper-left four grid locations in both E=0 and E=1 maps, then E is a don't-care. If you circle the quadrant in just the E=0 map, then the term you are circling must have E in the intersection.