I am a beginner. So, please correct me if this is wrong or sub-optimal.

internalSplit :: [a] -> Int -> [a] -> [[a]]
split :: [a] -> [[a]]

internalSplit (first:rest) count firstPart
    | count == 0 = [firstPart, (first:rest)]
    | otherwise  = internalSplit rest (count - 1) (firstPart ++ [first])

split myList =
    let listLength = length myList
    in
        if listLength `mod` 2 == 0 then
            internalSplit myList (listLength `div` 2) []
        else
            internalSplit myList ((listLength `div` 2) + 1) []

main = do
        print $ split [1, 2, 3, 5, 6]
        print $ split [1, 2, 3, 4, 5, 6]

Output

[[1,2,3],[5,6]]
[[1,2,3],[4,5,6]]

Edit:

Managed to use builtin functions and came up with this

internalSplit :: [a] -> Int -> [[a]]
split :: [a] -> [[a]]

internalSplit myList splitLength = [(take splitLength myList), (drop splitLength myList)]

split myList =
    let listLength = length myList
    in
        if listLength `mod` 2 == 0 then
            internalSplit myList (listLength `div` 2)
        else
            internalSplit myList ((listLength `div` 2) + 1)

main = do
        print $ split [1, 2, 3, 5, 6]
        print $ split [1, 2, 3, 4, 5, 6]

Output

[[1,2,3],[5,6]]
[[1,2,3],[4,5,6]]

Edit 1:

internalSplit :: [a] -> Int -> ([a], [a])
split :: [a] -> ([a], [a])

internalSplit myList splitLength = splitAt splitLength myList

split myList =
    let listLength = length myList
    in
        if listLength `mod` 2 == 0 then
            internalSplit myList (listLength `div` 2)
        else
            internalSplit myList ((listLength `div` 2) + 1)

main = do
        print $ split [1, 2, 3, 5, 6]
        print $ split [1, 2, 3, 4, 5, 6]

Output

([1,2,3],[5,6])
([1,2,3],[4,5,6])

Edit2

As suggested by Bogdon in the comments section, this can be greatly simplified to this

split :: [a] -> ([a], [a])
split myList = splitAt (((length myList) + 1) `div` 2) myList
main = do
        print $ split [1, 2, 3, 5, 6]
        print $ split [1, 2, 3, 4, 5, 6]

Output

([1,2,3],[5,6])
([1,2,3],[4,5,6])

There are Data.List.Split.chunksOf and Data.List.Grouping.splitEvery implementations of this routine in specialized packages (and a number included in other application packages: search by Int -> [a] -> [[a]] signature on Hayoo).

I think splitEvery implementation is pretty elegant:

splitEvery :: Int -> [a] -> [[a]]
splitEvery _ [] = []
splitEvery n xs = as : splitEvery n bs 
  where (as,bs) = splitAt n xs

First of all, note that the function you are trying to construct is already in the standard library, in the Prelude - it is called splitAt. Now, directly looking at its definition is confusing, as there are two algorithms, one which doesn't use the standard recursive structure at all -splitAt n xs = (take n xs, drop n xs) - and one that is hand-optimized making it ugly. The former makes more intuitive sense, as you are simply taking a prefix and a suffix and putting them in a pair. However, the latter teaches more, and has this overall structure:

splitAt :: Int -> [a] -> ([a], [a])
splitAt 0 xs     = ([], xs)
splitAt _ []     = ([], [])
splitAt n (x:xs) = (x:xs', xs'')
  where
    (xs', xs'') = splitAt (n - 1) xs

The basic idea is that if a list is made up of a head and a tail (it is of the form x:xs), then the list going from index k+1 onwards will be the same as the list going from k onwards once you remove the first element - drop (k + 1) (x : xs) == drop k xs. To construct the prefix, you similarly remove the first element, take a smaller prefix, and stick the element back on - take (k + 1) (x : xs) == x : take k xs.

Here's a maximally lazy approach:

splitWhen :: (a -> a -> Bool) -> [a] -> [[a]]
splitWhen f = foldr go [[]] where
  go x acc = (x:xs):xss where
    xs:xss = case acc of
      (z:_):_ | f x z -> []:acc
      _ -> acc

splitAtDup :: Eq a => [a] -> [[a]]
splitAtDup = splitWhen (==)

To verify the laziness, try this:

take 2 $ take 4 <$> splitAtDup (1:2:3:3:4:5:6:undefined)

It can be fully evaluated to normal form as [[1,2,3],[3,4,5,6]].

In a real program you should probably use

splitlist :: [a] -> ([a], [a])
splitlist xs = splitAt ((length xs + 1) `div` 2) xs

(i.e. something along the lines of dreamcrash's answer.)

But if, for learning purposes, you're looking for an explicitly recursive solution, study this:

splitlist :: [a] -> ([a], [a])
splitlist xs = f xs xs where
    f (y : ys) (_ : _ : zs) =
        let (as, bs) = f ys zs
        in (y : as, bs)
    f (y : ys) (_ : []) = (y : [], ys)
    f ys [] = ([], ys)