a)For critical points, we just set $x'=0$ and $y'=0$ which will yield four critical points: $P1=(0,0), P2=(-1,-1), P3=(2,2), P4=(-2,0)$. For linearization, we have to compute the Jacobian i.e the matrix of first derivatives of the functions. If $F=(2+x)(y-x), G=y(2+x-x^2)$, then $F_x=y-2x-2, F_y=2+x, G_x=y(1-2x), G_y=2+x-x^2$. Now we have got everything to find corresponding linearized systems, just use equation 13 from page 522 (book).

b) To determine the nature of solutions at each critical point, we just have to find the eigenvalues of the matrices we get from evaluating the Jacobian at the respective critical points.

For the point $P1$, we evaluate the Jacobian to find that the resulting matrix has eigenvalues $r_1=2, r_2=-2$, so the the critical point is a saddle (look picture from handout), this is unstable.

For the point $P2$, we evaluate the Jacobian to find that the resulting matrix has complex eigenvalues (the characteristic equation is $r^2+r+3=0$), so the the critical point is a spiral point, this is (asymptotically) stable as the real part of the eigenvalues is negative.

For the point $P3$, we evaluate the Jacobian to find that the resulting matrix has complex eigenvalues (the characteristic equation is $r^2+4r+24=0$), so the the critical point is a spiral point, this is (asymptotically) stable as the real part of the eigenvalues is negative.

For the point $P4$, we evaluate the Jacobian to find that the resulting matrix has eigenvalues $r_1=2, r_2=-4$, so the the critical point is a saddle, this is unstable.

For visualization of solutions, go to the course homepage and look at the section of learning resources. Pick the link to math.rice, select PPLANE.2005.10, type in the equations and enjoy the picture! You can also see the answer to the bonus question from the picture.