jmc

By the remainder theorem, $f(x)-f(a)$ is divisible by $x-a,$ so we can take out a factor of $x-a$ accordingly: $$\begin{array}{rl}f(x)-f(a)& =(x^3+3x^2+1)-(a^3+3a^2+1)\\ & =(x^3-a^3)+3(x^2-a^2)\\ & =(x-a)(x^2+ax+a^2)+3(x-a)(x+a)\\ & =(x-a)(x^2+ax+a^2+3x+3a)\\ & =(x-a)(x^2+(a+3)x+a^2+3a).\end{array}$$Thus, we want $$x^2+(a+3)x+a^2+3a=(x-a)(x-b)=x^2-(a+b)x+ab.$$Matching coefficients, we get $$\begin{array}{rl}a+3& =-a-b,\\ a^2+3a& =ab.\end{array}$$Since $a\ne 0,$ we can divide both sides of the second equation by $a,$ to get $a+3=b.$ Then $-a-b=b,$ so $a=-2b.$ Then $$-2b+3=2b-b,$$which gives us $b=1$. Then $a=-2,$ so $(a,b)=\boxed{(-2,1)}.$