If we solve the first system, we get
[tex]\begin{cases}y=x+1 \\ y=3x+3\end{cases}[/tex]Let's multiply the first expression by -1.
[tex]\begin{gathered} \begin{cases}-y=-x-1 \\ y=3x+3\end{cases}\rightarrow0=2x+2 \\ 2x=-2 \\ x=-\frac{2}{2} \\ x=-1 \end{gathered}[/tex]Hence, the first system has 1 solution.
Let's solve the second system
[tex]\begin{cases}3x+6y=-6 \\ 9x+18y=-18\end{cases}[/tex]As you can observe, the second equation is triple the first one, which means the equations are the same. In other words, the system has infinitely many solutions.
[tex]\begin{cases}y=4x-2 \\ -4x+y=5\end{cases}[/tex]Let's multiply the first equation by -1
[tex]\begin{gathered} \begin{cases}-y=-4x+2 \\ -4x+y=5\end{cases}\rightarrow-4x=-4x+2+5 \\ -4x+4x=7 \\ 0=7 \end{gathered}[/tex]Given that this result is false, we conclude that the system has no solutions.
