a) By translation we may assume $g(0)=0$. Fix $\epsilon>0$ and choose $\delta>0$ such that $|g(x)|<\epsilon$ for $|x|<\delta$ and such that $f$ is bounded on $(-\delta,\delta)$. Choose some $M$ such that $|f(x)|<M$ on $(-\delta,\delta)$. We define $c_0:=M, c_{n+1}:={{c_n+\epsilon}\over 2}$. Suppose that $|f(x)|\le c_n$ on $(-\delta,\delta)$ which is true for $n=0$. For $|x|<\delta$ we have $|2f(x)+f(x^2)|=|g(x)|\le \epsilon\implies 2|f(x)|\le \epsilon+c_n\implies |f(x)|\le c_{n+1}$. Hence $|f(x)|\le c_n$ on $(-\delta,\delta)$ for all $n$. If $n\rightarrow \infty$ then $c_n\rightarrow \epsilon$ since $|c_{n+1}-\epsilon|=1/2|c_n-\epsilon|$. Hence $|f(x)|\le \epsilon$ on $(-\delta,\delta)$. Hence $f$ is continuous in $x=0$. b) Let $\phi:[2,4)\rightarrow R$ be any function which is not $\equiv 0$. Let $h(x)=x^2$. If $x>0$ then we can write $x=h^n(\xi)$ where $n\in Z, \xi\in [2,4)$ and this notation is unique. Define $f(x)=(-2)^n \phi(\xi)$. We set $f(0)=0$ and we extend $f$ to an even function on $R$. Then $f$ is not bounded near $x=0$ hence not continuous. Also $g(x)=f(x^2)+2f(x)\equiv 0$ which is continuous.