"Does newton's method converge to root $\alpha \in [\frac\pi2, \pi]$ if the initial approximation is taken to be $x_0 = \pi$?"
I don't know how to deal with the lambda..
So Newton's Method is:
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$$
So, $f(x_n)$ will be $f(\pi) = \tan(\pi) + \lambda \pi = 0 + \lambda \pi$
$f'(\pi) = \sec^2(\pi) + \lambda = 1 + \lambda$
$x_{2} = \pi - \frac{\lambda \pi}{1+\lambda}$
but then how am I to conclude that this approaches the root $\alpha \in [\frac\pi2,\pi]$?
Starting with $x_0=\pi$ you obtain $$x_1={\pi\over 1+\lambda}\ .$$ When $\lambda=0$ then $x_0=\pi$ is already the solution, and when $\lambda=1$ then $\tan x_1$ is undefined, and the procedure immediately comes to an end.
When $0<\lambda<1$ then ${\pi\over2}<x_1<\pi$. The function $$f(x):=\tan x+\lambda x$$ is concave in $\ \bigl]{\pi\over2},\pi\bigr]\ $, and this implies that $f(x_1)<f(\pi)+(x_1-\pi)f'(\pi)=0$. The "upward convexity" property of $f$ now guarantees that starting with $x_1$ the $x_n$ $\>(n\geq1)$ found by Newton's procedure will increase and quadratically converge to the zero $\xi$ you are interested in.