Saving 99 of 100 prisoners with hat parity

One hundred prisoners stand in a line, each wearing a red or blue hat. Every prisoner can see all hats except their own. In order, each one calls out a guess of their own hat colour, and everyone hears every guess. Those who guess right are freed. They may agree on a plan beforehand. What strategy saves the most prisoners for certain? How does it extend if hats can be one of three colours?

$99$ are guaranteed in both cases. The first speaker announces the parity of the red hats they see (e.g. "red" for odd); each later prisoner compares that with the reds they see and hear named, and deduces their own colour. With three colours, code them $0,1,2$ and have the first speaker announce the sum of the hats they see modulo $3$.