Statement -1 : If $x < \sqrt{e}$, then 

$cot^{-1}\begin{Bmatrix}\frac{log(e/x^2)}{log(ex^2)}\end{Bmatrix}+cot^{-1}\begin{Bmatrix}\frac{log(ex^4)}{log(e^2/x^2)}\end{Bmatrix}= \pi - tan^{-1} 3 $

Statement-2: $tan^{-1}\left(\frac{x+y}{1-xy}\right) = tan^{-1}x + tan^{-1} y. $ if xy <1.

Statement 1 is True, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

Statement-2 is true (see theory).

Now,

$cot^{-1}\begin{Bmatrix}\frac{log(e/x^2)}{log(ex^2)}\end{Bmatrix}+cot^{-1}\begin{Bmatrix}\frac{log(ex^4)}{log(e^2/x^2)}\end{Bmatrix}$

$= cot^{-1}\begin{Bmatrix}\frac{1-2log x}{1+2log x}\end{Bmatrix}+cot^{-1}\begin{Bmatrix}\frac{1+4logx}{2-2log x}\end{Bmatrix}$

$=tan^{-1}\left(\frac{1+2log x}{1-2log x}\right)+cot^{-1}\left(\frac{\frac{1}{2}+2log x}{1-\frac{1}{2}×2log x}\right)$

$=tan^{-1}\left(\frac{1+2log x}{1-2log x}\right)+\frac{\pi}{2} - tan^{-1}\left(\frac{\frac{1}{2}+2log x}{1-\frac{1}{2}×2log x}\right)$

$= tan^{-1} (1) +tan^{-1}(2 log x) +\frac{\pi}{2} - tan^{-1}\frac{1}{2} - tan^{-1} (2 log x )$ if 2 log x < 1 and log x < 1

$=\frac{3\pi}{4} - tan^{-1}\frac{1}{2}$, if  $ x < \sqrt{e}$

$= \pi - (tan^{-1} 1 + tan^{-1}\frac{1}{2}) = \pi - tan^{-1} 3 $

So, statement -1 is also true. Also, statement-2 is a correct explanation for statement-1.