(a) Write \(f(x)=A/(1+2x)+B/(3-x)+C/(3-x)^2\). Multiplying by the denominator, values \(x=-1/2,3\) give \(A=2,C=-3\). Comparing \(x^2\) gives \(A-2B=-2\), so \(B=2\). Hence \(\boxed{f(x)=\dfrac2{1+2x}+\dfrac2{3-x}-\dfrac3{(3-x)^2}}\).

(b) The expansions are \(2(1-2x+4x^2)+\tfrac23(1+\tfrac x3+\tfrac{x^2}9)-\tfrac13(1+\tfrac{2x}3+\tfrac{x^2}3)+O(x^3)\). Collecting gives \(\boxed{f(x)=\tfrac73-4x+\tfrac{215}{27}x^2+O(x^3)}\).