Better typings for Array.concat()

[jablko]

Fixes #26378
Fixes #26976

[Jessidhia]

You probably need to keep at least the less-generic overload around as the first overload.

As it is, the free U parameter will prevent inference from working in cases where one is trying to build a homogeneous array; especially when the array includes function or object types that might have methods.

const foo = [(a: string) => a]
const bar = foo.concat(b => b) // implicit any error

[jablko]

@Jessidhia Thank you for this feedback! I've now updated the PR accordingly.

[RyanCavanaugh]

Rolled into #33645