(i)  strange f g = g (f g)

Assume g :: a -> b.  Then f :: (a -> b) -> c.  But since g :: a -> b,

f g :: a, so c = a.  Therefore, f :: (a -> b) -> a, and g (f g) :: a.

Therefore, strange :: ((a -> b) -> a) -> (a -> b) -> a.

(ii)  stranger f = f f

Assume f :: a -> b.  Since "f f" is well-typed, type unification requires

a = b.  Therefore, f :: a -> a, and stranger :: (a -> a) -> a.