jmc

Setting $x=-2$ in the equation $Q_1(x)(x+2)-13=Q_3(x)(x+2)(x^2-3x-4)+R(x),$ we get $$R(-2)=-13.$$Setting $x=4$ and $x=-1$ in the equation $Q_2(x)(x^2-3x-4)-5x-11=Q_3(x)(x+2)(x^2-3x-4)+R(x),$ we get $$R(4)=-31\text{and}R(-1)=-6.$$Since $\mathrm{deg}R(x)=2,$ we can let $R(x)=a{x}^2+bx+c.$ Then $$\begin{array}{rl}4a-2b+c& =-13,\\ 16a+4b+c& =-31,\\ a-b+c& =-6.\end{array}$$Subtracting these equations in pairs, we get $$\begin{array}{rl}12a+6b& =-18,\\ 3a-b& =-7.\end{array}$$Solving, we find $a=-2$ and $b=1,$ so $c=-3.$ Hence, $R(x)=\boxed{-2x^2+x-3}.$