Explanation

(a)
Given equation is compared with the standard form, we get
$a = 1,\ h = - \frac{3}{2},\ b = \lambda,\ g = \frac{3}{2},\ f = \frac{- 5}{2},\ c = 2$
$Given\ that,\ \theta = \tan^{- 1}\left( \frac{1}{3} \right) \Rightarrow \tan\theta = \frac{1}{3}$
$Since,\tan\theta = \frac{2\sqrt{h^{2} - ab}}{a + b}$
$\Rightarrow \frac{1}{3} = \frac{2\sqrt{\left( - \frac{3}{2} \right)^{2} - \lambda}}{\lambda + 1} \Rightarrow ({\lambda + 1)}^{2} = 9(9 - 4\lambda)$
$\Rightarrow \ \lambda^{2} + 1 + 2\lambda = 81 - 36\lambda$
$\Rightarrow \ \lambda^{2} + 38\lambda - 80 = 0$
$\Rightarrow \ \lambda = \frac{- 38 \pm \sqrt{(38)^{2} + 320}}{2}$
$\Rightarrow \ \lambda = \frac{- 38 \pm 42}{2}\ \Rightarrow \ \lambda = 2$