Consider Abel and Betty as a single unit or block. This reduces the problem to arranging 6 units: (AB), Cally, Doug, Eve, Freya, and Gino.

The number of ways to arrange these 6 units is given by:

\(6! = 720\)

Within the block (AB), Abel and Betty can be arranged in:

\(2! = 2\)

Thus, the total number of arrangements with Abel and Betty together is:

\(6! \times 2! = 720 \times 2 = 1440\)

Now, consider the arrangements where Freya and Gino are also together. Treat Freya and Gino as another block, reducing the problem to arranging 5 units: (AB), (FG), Cally, Doug, and Eve.

The number of ways to arrange these 5 units is:

\(5! = 120\)

Within the block (FG), Freya and Gino can be arranged in:

\(2! = 2\)

Thus, the total number of arrangements with both (AB) and (FG) together is:

\(5! \times 2! \times 2! = 120 \times 2 \times 2 = 480\)

Therefore, the number of arrangements where Abel and Betty are together but Freya and Gino are not together is:

\(1440 - 480 = 960\)