Given,

$f\left(x\right)=\frac{\left(\tan 1^{°}\right)x+{\log }_e\left(123\right)}{x {\log }_e\left(1234\right)-\left(\tan 1^{°}\right)}, x>0$

Now, let $\tan 1^{°}=a, {\log }_e\left(123\right)=b$ and ${\log }_e\left(1234\right)=c$

So, $f\left(x\right)=\frac{ax+b}{cx-a}$

$\Rightarrow f\left(f\left(x\right)\right)=\frac{af\left(x\right)+b}{cf\left(x\right)-a}$

$\Rightarrow f\left(f\left(x\right)\right)=\frac{a\left(\frac{ax+b}{cx-a}\right)+b}{c\left(\frac{ax+b}{cx-a}\right)-a}$

$\Rightarrow f\left(f\left(x\right)\right)=\frac{a^{2}x+ab+b\left(cx-a\right)}{acx+bc-a\left(cx-a\right)}=\frac{a^{2}x+bcx}{bc+a^2}=x$

$\Rightarrow f\left(f\left(x\right)\right)+f\left(f\left(\frac{4}{x}\right)\right)=x+\frac{4}{x}$

$\because x>0$, then least value $=\sqrt{x·\frac{4}{x}}=2$