(a)The critical points consist of the solution set of the equations. $$(2+x)(y-x) = 0$$ $$(4-x)(y+x) = 0$$, the only critical points are at (0,0), (4,4)and (-2,2).

(b,c) First note that $F(x) = (2+x)(y-x) $ ,$G(x,y) = (4-x)(y+x)$ . The Jacobian matrix of the vector field is $$J = \begin{pmatrix} -2-2x+y & 2+x \\ 4-2y-2x & 4-x \end{pmatrix}$$

At the origin, the coefficient matrix of the linearized system is $$J(0,0) = \begin{pmatrix} -2 & 2 \\ 4 & 4 \end{pmatrix}$$

with eigenvalues $r_1 = 1-\sqrt{17}$, $r_2 = 1+\sqrt{17}$ . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium point (-2,2), the coefficient matrix of the linearized system is $$J(-2,2) = \begin{pmatrix} 4 & 0 \\ 6 & 6 \end{pmatrix}$$

with eigenvalues $r_1 = 4$ and $r_2 = 6$ . The eigenvalues are real, unequal and positive, hence the critical point is an unstable node. At the point (4,4), the coefficient matrix of the linearized system is $$J(4,4) = \begin{pmatrix} -6 & 6 \\ -8 & 0 \end{pmatrix}$$

with complex conjugate eigenvalues $r_1 = -3 + \sqrt{39}i$, $r_2 = -3 - \sqrt{39}i$, The critical point is a stable spiral, which is asymptotically stable.

Attached is the part (d)