(i) (f + g)oh = foh + goh

Let us consider ((f + g)oh)(x) = (f + g)(h(x))

= f(h(x)) + g(h(x))

= (foh)(x) + (goh)(x)

= {(fog) + (goh)}(x)

Then, ((f + g)oh)(x) = {(foh) +(goh)}(x) ∀ x ϵ R

Therefore, (f + g)oh = foh + goh.

(ii) (f.g)oh = (foh).(goh)

Let us consider ((f.g)oh)(x) = (f.g)(h(x))

= f(h(x)).g(h(x))

= f (h(x)).g(h(x))

= (fog)(x).(goh)(x)

= {(fog).(goh)}(x)

Then, ((f.g)oh)(x) = {(fog).(goh)}(x) ∀ x ϵ R

Therefore, (f.g)oh = (fog).(goh)