(a) The range of g is negative, so its values are in the domain of f. \[fg(x)=\frac1{(2x+1)^2}-1.\] Setting this equal to 3 gives \[(2x+1)^2=\frac14,\qquad 2x+1=\pm\frac12.\] The roots are \(-1/4\) and \(-3/4\); only the latter is in \(x<-1/2\). Therefore \[\boxed{x=-3/4}.\]

(b) Put \(y=1/(2x+1)^2-1\). Then \[(2x+1)^2=\frac1{y+1}.\] Since \(2x+1<0\), choose the negative square root: \[2x+1=-\frac1{\sqrt{y+1}},\qquad x=-\frac12-\frac1{2\sqrt{y+1}}.\] Hence \[\boxed{(fg)^{-1}(x)=-\frac12-\frac1{2\sqrt{x+1}},\quad x>-1}.\]