Step-by-step Solution:

Given \[ x+y=1,\qquad x^2+y^2=2,\qquad x^5+y^5=A. \] Let \(s=x+y\) and \(p=xy\). Then \[ s=1,\qquad x^2+y^2=s^2-2p=1-2p=2. \] Hence \[ 1-2p=2\implies -2p=1\implies p=-\tfrac{1}{2}. \] The numbers \(x,y\) are the roots of \(t^2-st+p=t^2-t-\tfrac12=0\), whose discriminant is \[ \Delta=1-4\left(-\tfrac12\right)=1+2=3, \] so the two roots are distinct real numbers \[ x,\;y=\frac{1\pm\sqrt{3}}{2}. \] Therefore the ordered solution pairs \((x,y)\) are \(\big(\tfrac{1+\sqrt3}{2},\tfrac{1-\sqrt3}{2}\big)\) and its swap, so the number of solution pairs is \[ N=2. \] To find \(A=x^5+y^5\) use the recurrence for power sums for two variables: \[ S_n:=x^n+y^n,\qquad S_n=sS_{n-1}-pS_{n-2}. \] Here \(s=1,\;p=-\tfrac12\), so \[ S_n=S_{n-1}+\tfrac12S_{n-2}. \] We have \(S_0=2,\;S_1=s=1,\;S_2=2\). Then \[ S_3=S_2+\tfrac12S_1=2+\tfrac12=\tfrac{5}{2}, \] \[ S_4=S_3+\tfrac12S_2=\tfrac{5}{2}+1=\tfrac{7}{2}, \] \[ S_5=S_4+\tfrac12S_3=\tfrac{7}{2}+\tfrac12\cdot\tfrac{5}{2}=\tfrac{7}{2}+\tfrac{5}{4}=\tfrac{19}{4}. \] Thus \(A=S_5=\dfrac{19}{4}\). Finally \[ AN=\frac{19}{4}\cdot 2=\frac{19}{2}. \] \[ \boxed{AN=\dfrac{19}{2}} \]