Again use structural induction on the finite list. At [], both the right fold and the left fold return e. For x:xs, use reverse (x:xs) = reverse xs ++ [x], part (a), and the definition of flip:
foldl (flip f) e (reverse (x:xs))
= foldl (flip f) e (reverse xs ++ [x])
= foldl (flip f) (foldl (flip f) e (reverse xs)) [x]
= flip f (foldl (flip f) e (reverse xs)) x
= f x (foldl (flip f) e (reverse xs))
= f x (foldr f e xs)
= foldr f e (x:xs)
The penultimate equality uses the induction hypothesis. Thus foldr f e xs = foldl (flip f) e (reverse xs). Finiteness matters: reversing an infinite list does not produce its first element, whereas some right folds do.

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