(*) Split a list into two parts; the length of the first part is given.
Do not use any predefined predicates.
split xs n = (take n xs, drop n xs)
Or even simpler using
split = flip splitAt
But these should clearly be considered "predefined predicates". Alternatively, we have the following recursive solution:
split :: [a] -> Int -> ([a], [a]) split  _ = (, ) split l@(x : xs) n | n > 0 = (x : ys, zs) | otherwise = (, l) where (ys,zs) = split xs (n - 1)
The same solution as above written more cleanly:
split :: [a] -> Int -> ([a], [a]) split xs 0 = (, xs) split (x:xs) n = let (f,l) = split xs (n-1) in (x : f, l)
A similar solution using foldl:
split :: [a] -> Int -> ([a], [a]) split  _ = (, ) split list n | n < 0 = (list, ) | otherwise = (first output, second output) where output = foldl (\acc e -> if third acc > 0 then (first acc ++ [e], second acc, third acc - 1) else (first acc, second acc ++ [e], third acc)) (, , n) list
Note that for the above code to work you must define your own first, second, and third functions for tuples containing three elements like so:
first :: (a, b, c) -> a first (x, _, _) = x second :: (a, b, c) -> b second (_, y, _) = y third :: (a, b, c) -> c third (_, _, z) = z
Another foldl solution without defining tuple extractors:
split :: [a] -> Int -> ([a],[a]) split lst n = snd $ foldl helper (0,(,)) lst where helper (i,(left,right)) x = if i >= n then (i+1,(left,right++[x])) else (i+1,(left++[x],right))