*Main> :t (.)
(.) :: (b -> c) -> (a -> b) -> a -> c

Write as prefix instead of infix

(.) . (.) <=> (.) (.) (.)
    ^          ^

In the same way as
1 + 2 <=> (+) 1 2

Now, rename the types in the second and third (.)
For example:
    For the second (.):
        (.) :: (x2 -> x3) -> (x1 -> x2) -> x1 -> x3
    And the third (.):
        (.) :: (y2 -> y3) -> (y1 -> y2) -> y1 -> y3

Now, the first argument (b -> c) of the first (.) is given the second (.). This means that:
    (b -> c) = (x2 -> x3) -> (x1 -> x2) -> x1 -> x3
        and therefore
            b: (x2 -> x3)
            c: (x1 -> x2) -> x1 -> x3

And now the same for the second argument (a -> b) that's given the third (.):
    (a -> b) = (y2 -> y3) -> (y1 -> y2) -> y1 -> y3
        and therefore:
            a: (y2 -> y3)
            b: (y1 -> y2) -> y1 -> y3

Solving the type of b:
    (x2 -> x3) = (y1 -> y2) -> y1 -> y3
        therefore:
            x2 = (y1 -> y2)
            x3 = y1 -> y3

Knowing this, we can obtain the type of c:
    c = (x1 -> x2) -> x1 -> x3
      = (x1 -> y1 -> y2) -> x1 -> y1 -> y3

And a:
    a = (y2 -> y3)

Going back to our whole expression:
(.).(.) = (.) (.) (.) :: a -> c
    = (y2 -> y3) -> (x1 -> y1 -> y2) -> x1 -> y1 -> y3

And we can rename the types as in the solutions:
    (b -> c) -> (a1 -> a -> b) -> a1 -> a -> c