The line y=1 is a horizontal line paralel to the x axis. This means that a reflection about it can't change the value of the x coordinate:
![A_x=3=A^(\prime)_x](https://img.qammunity.org/2023/formulas/mathematics/college/g9in0ipo4498urb46h59st86gzrpn73t50.png)
Regarding the y coordinate, in A we have that:
![\begin{gathered} A_y=-2 \\ y-A_y=1-(-2)=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/sh0q9vi42xecaj8x94rwq5h4ubb10w4cip.png)
Since A' is the reflection of A about line y=1 then we have that:
![\begin{gathered} y-A^(\prime)_y=-(y-A^{}_y))=-3 \\ y-A^(\prime)_y=1-A^(\prime)_y=-3 \\ A^(\prime)_y=1+3=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yayracdpj497609u1vn5bcha25i1ysuu2w.png)
Then the coordinates of point A' are (3,4) and the correct answer is the second one.