:: a -> (a -> b) -> b -package:diagrams-lib -package:quaalude -package:lens-family-core -package:relude package:protolude

& is a reverse application operator. This provides notational convenience. Its precedence is one higher than that of the forward application operator $, which allows & to be nested in $.
>>> 5 & (+1) & show
"6"
($) is the function application operator. Applying ($) to a function f and an argument x gives the same result as applying f to x directly. The definition is akin to this:
($) :: (a -> b) -> a -> b
($) f x = f x
On the face of it, this may appear pointless! But it's actually one of the most useful and important operators in Haskell. The order of operations is very different between ($) and normal function application. Normal function application has precedence 10 - higher than any operator - and associates to the left. So these two definitions are equivalent:
expr = min 5 1 + 5
expr = ((min 5) 1) + 5
($) has precedence 0 (the lowest) and associates to the right, so these are equivalent:
expr = min 5 $ 1 + 5
expr = (min 5) (1 + 5)

Uses

A common use cases of ($) is to avoid parentheses in complex expressions. For example, instead of using nested parentheses in the following Haskell function:
-- | Sum numbers in a string: strSum "100  5 -7" == 98
strSum :: String -> Int
strSum s = sum (mapMaybe readMaybe (words s))
we can deploy the function application operator:
-- | Sum numbers in a string: strSum "100  5 -7" == 98
strSum :: String -> Int
strSum s = sum $ mapMaybe readMaybe $ words s
($) is also used as a section (a partially applied operator), in order to indicate that we wish to apply some yet-unspecified function to a given value. For example, to apply the argument 5 to a list of functions:
applyFive :: [Int]
applyFive = map ($ 5) [(+1), (2^)]
>>> [6, 32]

Technical Remark (Representation Polymorphism)

($) is fully representation-polymorphic. This allows it to also be used with arguments of unlifted and even unboxed kinds, such as unboxed integers:
fastMod :: Int -> Int -> Int
fastMod (I# x) (I# m) = I# $ remInt# x m