Answer
$$[x]_B=\left[ \frac{4\sqrt{13}}{13} \quad \frac{7\sqrt{13}}{13}\right]^T.$$
Work Step by Step
Let $x=(1,2)$ and $B=\{ (-\frac{2\sqrt{13}}{13},\frac{3\sqrt{13}}{13}),(\frac{3\sqrt{13}}{13},\frac{2\sqrt{13}}{13}) \}$
To find the coordinates of $x$ relative to $B$, we have to find the following;
$$\langle (1,2),(-\frac{2\sqrt{13}}{13},\frac{3\sqrt{13}}{13})\rangle =-\frac{2\sqrt{13}}{13}+\frac{6\sqrt{13}}{13}=\frac{4\sqrt{13}}{13}$$
$$\langle (1,2),(\frac{3\sqrt{13}}{13},\frac{2\sqrt{13}}{13})\rangle =\frac{3\sqrt{13}}{13}+\frac{4\sqrt{13}}{13} =\frac{7\sqrt{13}}{13}.$$
Then, the coordinate matrix of $x$ relative to $B$ is given by
$$[x]_B=\left[ \frac{4\sqrt{13}}{13} \quad \frac{7\sqrt{13}}{13}\right]^T.$$
