At a function

$\begin{aligned} 2 A(x) & = e^x - e^{-x} \\ \Rightarrow A(x) & = \frac{e^x - e^{-x}}{2} = \sinh x \\ A(B(x)) & = x \\ \Rightarrow \sinh \left(\sinh^{-1} x \right) & = x \\ \Rightarrow B(x) & = \sinh^{-1} x \\ \Rightarrow B \left(\frac{e^{2890}-1}{2e^{1445}} \right) & = \sinh^{-1} \left(\frac{e^{1445}-e^{-1445}}{2} \right) \\ & = \sinh^{-1} \left( \sinh 1445 \right) \\ & = \boxed{1445} \end{aligned}$

It is worthy to note that if $A(B(x))=x$ then, $B(x)=A^{-1}(x)$ , however this solution does not utilize this fact.

We can solve for $B(x)$ thusly:

$2A(B(x))=2x=e^{B(x)}-e^{-B(x)}\Rightarrow 2xe^{B(x)}=e^{2B(x)}-1$

$e^{2B(x)}-2xe^{B(x)}=1$

$e^{2B(x)}-2xe^{B(x)}+x^2=1+x^2$ .

Let $e^{B(x)}=u$ .

Then:

$u^2-2ux+x^2=1+x^2\Rightarrow (u-x)^2=1+x^2\Rightarrow u=x+\sqrt{1+x^2}$ .

Therefore:

$B(x)=\ln (x+\sqrt{1+x^2} )$ .

Now if we let $e^{1445}=z$ , then:

$B(\frac{e^{2890}-1}{2e^{1445}})=B(\frac{z^2-1}{2z})=\ln\left (\frac{z^2-1}{2z}+\sqrt{1+(\frac{z^2-1}{2z})^2}\right )$ .

Expansion and simplification under the square root yields:

$\ln\left (\frac{z^2-1}{2z}+\sqrt{\frac{z^4-2z^2+1+4z^2}{4z^2}}\right )=\ln\left (\frac{z^2-1}{2z}+\sqrt{\frac{z^4+2z^2+1}{4z^2}}\right )=\ln\left (\frac{z^2-1}{2z}+\sqrt{\frac{(z^2+1)^2}{4z^2}}\right )=\ln\left (\frac{z^2-1}{2z}+\frac{z^2+1}{2z}\right )$ .

Therefore:

$B(\frac{z^2-1}{2z})=\ln\left (\frac{2z^2}{2z}\right )=\ln(z)$ ,

and:

$B(\frac{e^{2890}-1}{2e^{1445}})=\ln(e^{1445})=\boxed{1445}$ .