Ta có:

a)

Có $90^{o} < \alpha < 180^{o}$ => $cosα< 0$ 

$sin^{2}\alpha + cos^{2}\alpha = 1 $

=>$ cos^{2}\alpha = 1 - sin^{2}\alpha = 1 - \frac{1}{3} = \frac{8}{9}$

=>$ cos\alpha = -\frac{2\sqrt{2}}{3}$

Áp dụng công thức:

$tan\alpha = \frac{sin\alpha}{cos\alpha} = \frac{\frac{1}{3}}{\frac{-2\sqrt{2}}{3}} = \frac{-\sqrt{2}}{4}$

$cot\alpha = \frac{cos\alpha}{sin\alpha} = \frac{\frac{-2\sqrt{2}}{3}}{\frac{1}{3}} = -2\sqrt{2}$

b) 

$cot(180^{o} - \alpha) = -cot\alpha$

$cos(180^{o} - \alpha) = -cos\alpha$

$cot(90^{o} - \alpha) = tan\alpha$

Áp dụng vào biểu thức $A = sin\alpha . cot(180^{o} - \alpha) + cos(180^{o} - \alpha) .cot(90^{o} - \alpha)$:

$A = sin\alpha . (-cot\alpha) + (-cos\alpha) . tan\alpha$

$A = \frac{1}{3} . (-(-2\sqrt{2})) + (-(-\frac{2\sqrt{2}}{3})).(\frac{-\sqrt{2}}{4})$

$A = \frac{2\sqrt{2} - 1}{3}$

$B = \frac{3(sin\alpha + \sqrt{2}cos\alpha)-2}{sin\alpha -\sqrt{2}cos\alpha}$

$B = \frac{3(\frac{1}{3} + \sqrt{2}.(-\frac{2\sqrt{2}}{3}))-2}{\frac{1}{3} -\sqrt{2}.(-\frac{2\sqrt{2}}{3})}$

$B = -3$