(a) The numerator and denominator have equal degree, so include a constant: \[f(x)=A+\frac B{x+3}+\frac{Cx+D}{x^2+2}.\] Multiplying by \((x+3)(x^2+2)\) gives \[x^3+2x-11=A(x+3)(x^2+2)+B(x^2+2)+(Cx+D)(x+3).\] Matching \(x^3\) gives \(A=1\). Substituting \(x=-3\) gives \(B=-4\). The remaining \(x^2\) and \(x\) coefficients give \(C=1,D=-3\). Thus \[f(x)=1-\frac4{x+3}+\frac{x-3}{x^2+2}.\]

(b) Expand each denominator: \[\frac1{x+3}=\frac13\left(1-\frac x3+\frac{x^2}{9}+\cdots\right),\quad\frac1{x^2+2}=\frac12\left(1-\frac{x^2}{2}+\cdots\right).\] Hence \[\begin{aligned}f(x)&=1-\frac43+\frac49x-\frac4{27}x^2+\frac x2-\frac32+\frac34x^2+\cdots\\&=-\frac{11}{6}+\frac{17}{18}x+\frac{65}{108}x^2+\cdots.\end{aligned}\] The common convergence range is \(|x|<\sqrt2\).