 Solution

Set both $\oplus$ and $\otimes$ equal to list concatenation, with $a=[]$. Associativity gives xs ++ (ys ++ zs) = (xs ++ ys) ++ zs, which is equation (1). The two identity laws give xs ++ [] = [] ++ xs, which is equation (2). Part (b) therefore yields

concat :: [[a]] -> [a]
concat = foldl (++) []

Thus $\boxed{\mathrm{concat}=\mathrm{foldl}(++)[]}$ on finite lists. The argument order is unchanged: the accumulator is the already concatenated prefix, and the next inner list is appended to its right. This preserves the order but repeatedly copies that prefix.