$\cos(2)$=$\dfrac{1}{3}$ $\Rightarrow$$\sin(2)$=$\sqrt{1-\cos^{2}2}$
= $\sqrt{1 - \dfrac{1}{9}}$=$\dfrac{2\sqrt{2}}{3}$
B = $\dfrac{\sin2 - 3\cos2}{\sin2 - 2\cos2}$
= $\dfrac{\dfrac{2\sqrt{2}}{3}-3.\dfrac{1}{3}}{\dfrac{2\sqrt{2}}{3}-2.\dfrac{1}{3}}$
= $\dfrac{7 - 5\sqrt{2}}{2}$