(a) Write \[f(x)=\frac{A}{3x+1}+\frac{Bx+C}{x^2+4}.\] Then \[36-4x-3x^2=A(x^2+4)+(Bx+C)(3x+1).\] Comparing coefficients gives \(A+3B=-3\), \(B+3C=-4\), \(4A+C=36\). Solving, \(A=9,B=-4,C=0\). Hence \[\boxed{f(x)=\frac9{3x+1}-\frac{4x}{x^2+4}}.\]

(b) \[\begin{aligned}f(x)&=9(1-3x+9x^2-27x^3+\cdots)-x\left(1-\frac{x^2}{4}+\cdots\right)\\&=\boxed{9-28x+81x^2-\frac{971}{4}x^3+\cdots}.\end{aligned}\]

(c) The two expansions require \(|3x|<1\) and \(|x^2/4|<1\), respectively. Their common range is \(\boxed{|x|<1/3}\).