1. The (==) function in the second equation of the Maybe instance of (==) is not complete yet, since you need to match on instances of the Maybe type and "Just :: a -> Maybe a". Your idea "Just a == Just a" goes in the right direction, but please note that you can't bind two variable names ("a") in the same equation. You need to give each "boxed value" a different name. I'm sure you can work it out from there.
2. The right hand side of "(x:xs) == (y:ys)" is not implicitly recursive, but rather explicitly! Since we apply the function (==) to the smaller lists "xs" and "ys" until we arrive at a base case.
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