For part(a)

Let

\begin{equation}

x(x-y+1)=0

\end{equation}

and

\begin{equation}

y(x-2)=0

\end{equation}

We will have all the three critical points

\begin{equation}

(x,y)=(0,0),(2,3) or (-1,0)

\end{equation}

For part(b)

\begin{equation}

F=x(x-y+1)

\end{equation}

\begin{equation}

G=y(x-2)

\end{equation}

Therefore, the Matrix $J$

\begin{equation}

J={

\left[\begin{array}{ccc}

2x-y+1 & -x \\

y & x-2

\end{array}

\right ]},

\end{equation}

At point(0,0), we have

\begin{equation}

J[0,0]={

\left[\begin{array}{ccc}

1 & 0 \\

0 & -2

\end{array}

\right ]},

\end{equation}

Eigenvalues are

\begin{equation}

\lambda_1=-2

\end{equation}

\begin{equation}

\lambda_2=1

\end{equation}

Therefore, (0,0) is a saddle point and thus unstable.

At point(2,3), we have

\begin{equation}

J[2,3]={

\left[\begin{array}{ccc}

2 & -2 \\

3 & 0

\end{array}

\right ]},

\end{equation}

\begin{equation}

\lambda_3=1+\sqrt{5}i

\end{equation}

\begin{equation}

\lambda_4=1-\sqrt{5}i

\end{equation}

Therefore, (2,3) is a spiral point and is unstable. The orientation is counterclockwise.

\begin{equation}

J[-1,0]={

\left[\begin{array}{ccc}

-1 & 1 \\

0 & -3

\end{array}

\right ]},

\end{equation}

\begin{equation}

\lambda_5=-1

\end{equation}

\begin{equation}

\lambda_6=-3

\end{equation}

Therefore, (-1,0) is a node and is asymptotically stable.