Abstract algebra rings question

http://math.berkeley.edu/~williams/113/hw8.pdf

Please do #3.

One of the distributive laws fails. In all the following, let f, g, h ∈ F and let x ∈ R.

Closure under addition: for all f, g, x we know (f+g)(x) = f(x) + g(x) is well-defined and is in R, so (f+g) is in F.
Associativity of addition: for all f, g, h, x we have
((f+g)+h)(x) = (f+g)(x) + h(x) = (f(x) + g(x)) + h(x) = f(x) + (g(x) + h(x)) = f(x) + (g+h)(x) = (f + (g+h))(x).
So (f+g)+h = f+(g+h).
Additive identity: Define z as z(x) = 0. Then for all x, (f+z)x = f(x) + z(x) = (z+f)(x) = f(x). So f+z = f = z+f.
Additive inverse: define (-f)(x) = -f(x). Then for all x, (f+(-f))(x) = f(x) + (-f)(x) = f(x) + (-f(x)) = 0 = z(x). So f + (-f) = z, similarly -f + f = z.
Commutativity of addition: for all x (f+g)(x) = f(x) + g(x) = g(x) + f(x) = (g+f)(x), so f + g = g + f.
Closure under multiplication: for all f, g, x we have (f◦g)(x) = f(g(x)); since f, g are in F this is a well-defined function of x and (f◦g)(x) ∈ R. So (f◦g) ∈ F.
Associativity of multiplication: for all f, g, h, x we have ((f◦g)◦h)(x) = (f◦g)(h(x)) = f(g(h(x))) = f(g◦h(x)) = (f◦(g◦h))(x). So (f◦g)◦h = f◦(g◦h).
Existence of unit: let j(x) = x. Then for all f in F, (f◦j) = f = (j◦f).
Distributive laws: for all f, g, h, x we have
((f+g)◦h)(x) = (f+g)(h(x)) = f(h(x)) + g(h(x)) = (f◦h)(x) + (g◦h)(x), so (f+g)◦h = (f◦h) + (g◦h).
However, the other distributive law fails: (f◦(g+h))(x) = f((g+h)(x)) = f(g(x) + h(x))
which in general will not be equal to f(g(x)) + f(h(x)). For instance, let f(x) = 1, then f◦(g+h)(x) = 1 but (f◦g)(x) + (f◦h)(x) = 2. So f ◦ (g+h) ≠ (f◦g) + (f◦h).