Is it possible to add two signed 8-bit numbers together and set both the carry and overflow bits?
Per your comments, your question seems to be "is it possible to have both carry and overflow set for a two's complement add involving signed number?" It is. The typical implementation is to take the exclusive-OR of the carry-in for the last adder with the carry-out at the end of the chain -- hence, an overflowing addition of negative numbers will cause the carry-out bit to be set and the overflow bit to be set.
Here's an example, add -1 to -128:
Carry 10000 0000
1000 0000 (-128)
1111 1111 (-1)
---------
0111 1111 (oops, this is 127!)
Carry will be set, since the last add resulted in a carry -- and overflow will be set based on the rule above (also, note that -128 added to -1 is obviously not 127)