Stage 1:

Proposition:

If all 2-cell cages have the same pair of numbers, then these numbers can't be consecutive.

Proof:

Suppose all 2-cell cages are {AB}. Then all 5-cell cages can't have {AB}.

(Otherwise sum for the 5-cell cage >= 1+2+3+4+A|B = 10+A|B > A+B.)

Now r12c9=r45c6=r56c1=r89c7={AB}

NP r45c6={AB} @ c6,n5 => r45c5+r89c6 can't be from {AB}

NP r1c9+r5c1={AB} @ r159c159 => r9c5 can't be from {AB}

HP @ c5: r38c5={AB} => NP r8c57={AB} @ r8 => r8c4 can't be from {AB}

=> Either r7c4+r8c5 or r8c5+r9c4={AB} @ n8

=> FNC: A & B can't be consecutive

Application:

The common cage sum can't be 17, as 2-cell cages can't be all {89}.

Therefore the common cage sum must be 15 or 16:

5-cell cage = {12345|12346}

2-cell cage = {69|78|79}

Stage 2:

Let the 5-cell cages be {PQRST}, and the other 4 candidates be {WXYZ}.

Then all 2-cell cages must be from {WXYZ}.

Now colour all known cells with {PQRST} in orange, and those with {WXYZ} in purple.

r12c5+r123c6, r2c8+r3c789+r4c9, r5c234+r6c45, r567c7+r7c56, r789c1+r8c23 are all orange

r12c9, r45c6, r56c1, r89c1 are all purple

=> n2,n7,r234c678: r123c4+r3c5, r79c23, r2c7+r4c78 are all purple

=> c7: r1c7 is orange => n3: r1c8 is purple

Now n8 must have 4 purple cells, with at most one @ r789c4 (c4) & at most one @ r9c456 (r9).

=> r8c56 are purple => r9c4 is orange, both r78c4 & r9c56 are one orange, one purple

=> r159c234: r1c23 are one orange, one purple => r1: r1c1 is orange

=> r678c159: r678c9 are one orange, two purple => c9: r59c9 are both orange

=> r159c159: r59c5 are both purple => c5: r4c5 is orange => r5: r5c8 is purple

=> r9: r9c68 are both orange => c6: r6c6 is purple => n5: r4c4 is orange

Stage 3:

Let C be the common cage sum.

Observe that if W+X=C, then all of W+Y, W+Z, X+Y, X+Z must not be C.

Let r12c9={WX} (without loss of generality), so W+X=C.

Now colour all known cells with {WX} in pink, and those with {YZ} in blue.

Suppose a 2-cell cage contains a pink cell, i.e. one cell is W|X.

Then the other cell = C-W|C-X = X|W, i.e. must be pink.

Thus if a 2-cell cage contains a pink cell, the other cell must also be pink.

r12c9 are both pink => n3: r2c8+r3c7 are both blue

=> c7: at least one of r89c7 must be pink => r89c7 are both pink

=> c7: r4c7 is blue => r234c678: r4c68 are both pink => r5c6 is pink

=> c6: r68c6 are both blue => n5: r5c5 is blue => r159c678: r5c8 is blue

=> r5: r5c1 is pink => r6c1 is pink => r159c159: r9c5 is blue => c5: r38c5 are both pink

Now n9 must have 2 blue cells, but they can't be @ r78c8 (c8).

=> r78c9 are both blue => c9: r6c9 is orange => n6: r6c8 is pink

=> c8: r78c8 are both orange => r8: r8c4 is orange => n8: r7c4 is pink

Observe that @ n8, there are 2 diagonally adjacent pink cells & 2 such blue cells.

=> {WX} & {YZ} both can't be {78} => C can't be 15

=> C=16 => {PQRST}={12346}, {WX}={79}, {YZ}={58}

Stage 4:

Apply FNC on all {79} cells: r59c5=r86c6=r42c7=r15c8=r78c9=[58]

Apply FNC on all {5} cells: r3c68, r46c4, r7c57, r8c8 all from {123}

r6c1 from {79} => FNC: r7c2 can't be 8 => r7: r7c3=8, r7c2 from {79}

r5c1 from {79} => FNC: r4c2 can't be 8 => r4: r4c1=8, r4c23 from {12346}

r7c7+r8c8 from {123} => FNC: r7c7+r8c8={13} (NP @ n9,r678c678)

=> FNC @ r8c8: r9c9 from {46} => 2 @ c8,n9 locked @ r79c8 => r38c8={13} (NP @ c8)

=> FNC @ r3c8: r4c9 can't be 2 => 2 @ r3,n3,5-cell cage locked @ r3c79

=> r3c68={13} (NP @ r3,r234c678) => r2c8+r3c79={246} (NT @ n3,5-cell cage)

=> FNC @ r1c7: r2c6 can't be 2 => r2c68={46} (NP @ r2,r234c678) => r3c7=2

=> c7: r56c7={46} (NP @ n6,5-cell cage) => r7c6=2, r7c57={13} (NP @ r7) => n9: r9c8=2

=> FNC @ r3c6: r24c5 can't be 2 => r2c5+r3c6={13} (NP @ n2) => n2: r1c5=2

=> c5: r46c5={46} (NP @ n5) => c4: r89c4={46} (NP @ n8) => r159c159: r5c9 can't be 2

=> n6: r6c9=2 => r234c159: r2c1=2 => c1: r3c1=5 => c4: r2c4=5 => FNC: r13c3 can't be {46}

=> r3: r3c2 from {46} => r3: r3c4=8 => r1c4, r3c3 from {79}, FNC: r2c3 can't be {79}

=> r2: r2c2 from {79} => r2: r2c3 from {13} => r1: r1c2=8 => r1: r1c3 from {13}

=> c2: r9c2=5 => r9c3 from {79} => c3: r6c3=5 => r6: r6c24={13} (NP @ r678c234)

=> FNC @ r9c2: r8c3=2 => c3: r45c3={46} (NP @ n4) => n9: r89c1={13} (NP @ c1)

4 @ r2 locked @ r2c68 => FNC: r1c7 can't be 3

The rest are singles under FNC.