$\because \; \; x+\frac{1}{x}=1\Rightarrow x^2-x+1=0$

$\Rightarrow$$\left(x+\omega \right)\left(x+{\omega }^2\right)=0$

$\therefore$ $x=-\omega ,-{\omega }^2$, where $\omega$ and ${\omega }^2$ are complex cube roots of unity

if $x=-\omega ,$  then  $\lambda ={\left(-\omega \right)}^{4000}+\frac{1}{{\left(-\omega \right)}^{4000}}={\omega }^{4000}+\frac{1}{{\omega }^{4000}}$

$\Rightarrow \lambda$ =  $\omega +\frac{1}{\omega }=\frac{{\omega }^2+1}{\omega }=\frac{-\omega }{\omega }=-1$     $\left(\begin{array}{c}\because {\omega }^3=1\\ \mathrm{and} 1+\omega +1{\omega }^2=0\end{array}\right)$

and also for  $x=-{\omega }^2\mathrm{, }\lambda =-1$

Now, for $n>1, 2^n=4k, k\in N$

$\therefore$ $2^{2^n}=2^{4k}={\left(16\right)}^k=$a number with last digit $6$

$\therefore$  $\mu$ = the digit at the unit place of the number $\left(2^{2^n}+1\right)$ 
i.e. $\mu =6+1=7$
Hence, $\lambda +\mu =-1+7=6$