
Since $A$ is the center of the dilation, it does not move and we will have $A^\prime = A$. For vertex $B$, we know that $B^\prime$ will lie on $\overleftrightarrow{AB}$ because dilations preserve lines through the center of dilation. Since the scale factor is $\frac{1}{3}$ we know that $AB^\prime = \frac{1}{3}AB$. From $A$ to $B$ is 4 units to the right and one unit up. Therefore, one third of the way from $A$ to $B$ (along $\overline{AB}$) will be the point $B^\prime = \left(1+ \frac{4}{3}, 1 + \frac{1}{3}\right)$. Applying this technique to $C$ which is 2 units to the right and 4 units up from $A$, we find $$ C^\prime = \left(1\frac{2}{3}, 2\frac{1}{3} \right). $$
The scaled triangle $A^\prime B^\prime C^\prime$ is pictured below:

Since $B$ is the center of the dilation, it does not move and we will have $B^{\prime \prime} = B$. For vertex $A$, we know that $A^{\prime \prime}$ will lie on $\overleftrightarrow{AB}$ because dilations preserve lines through the center of dilation. Since the scale factor is $\frac{2}{3}$ we know that $BA^{\prime \prime} = \frac{2}{3}BA$. From $B$ to $A$ is 4 units to the left and one unit down. Therefore, two thirds of the way from $B$ to $A$ (along $\overline{AB}$) will be the point $A^{\prime \prime} = \left(5  \frac{8}{3}, 2  \frac{2}{3}\right)$. Applying this technique to $C$ which is 2 units to the left and 3 units up from $B$, we find $$ C^{\prime \prime} = \left(3\frac{2}{3}, 4 \right). $$
The scaled triangle $A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$ is pictured below:

The calculations above show that $B^\prime = A^{\prime \prime} = \left( 2 \frac{1}{3}, 1 \frac{1}{3} \right)$. To see why these vertices are the same, note that the sum of the dilation factors $\frac{1}{3}$ and $\frac{2}{3}$ is 1. The dilation about $A$ maps $\overline{AB}$ to a the first third of $\overline{AB}$ while the dilation about $B$ maps $\overline{AB}$ to the second two thirds of this segment. The two scaled triangles with a shared vertex are pictured below: