Olá, Dan.
[tex](a)\text{ \underline{Contra-exemplo}: }f(x)=\log x, g(x) = h(x) = x \\\\ f \circ (g+h) = f(g(x)+h(x))=\log (x + x) = \log(2x) = \log 2 + \log x \\\\ f \circ g + f \circ h = f(g(x))+f(h(x))=\log x + \log x=2\log x \\\\ \Rightarrow \boxed{f \circ (g+h) \neq f \circ g + f \circ h}[/tex]
[tex](b)\text{ Seja }u(x)=f(x)+g(x) \Rightarrow u \circ h=f \circ h+g \circ h \Rightarrow \\\\ \begin{cases} f \circ h=u \circ h-g \circ h \\ g \circ h=u \circ h-f \circ h \end{cases} \\\\\\ \text{Somando as duas equa\c{c}\~oes, temos:} \\\\ f \circ h+g \circ h=2u \circ h-g \circ h-f \circ h \Rightarrow \\\\ 2f \circ h+2g \circ h =2u \circ h \Rightarrow \\\\ 2[f \circ h+g \circ h]=2u \circ h \Rightarrow \\\\ u \circ h=f \circ h+g \circ h \Rightarrow \\\\ \boxed{(f+g)\circ h=f \circ h+g \circ h}[/tex]
