Analysis

This problem examines the use of derivatives to study the maximum and minimum values of functions and the use of basic inequalities to find these values. By $f(x)-g(x)=x+e^{x-a}-\ln(x+2)+4e^{a-x}$, using derivatives and basic inequalities, it can be proven that $f(x)-g(x) \geqslant 3$ to solve the problem.

Solution

Given $f(x)-g(x)=x+e^{x-a}-\ln(x+2)+4e^{a-x}$,

Let $y=x-\ln(x+2)$,

$y' = 1 - \frac{1}{x+2} = \frac{x+1}{x+2}$,

Thus, $y=x-\ln(x+2)$ is a decreasing function on $(-2,-1)$ and an increasing function on $(-1,+\infty)$,

Therefore, when $x=-1$, $y$ has a minimum value of $-1-0=-1$,

By the basic inequality, we have $e^{x-a}+4e^{a-x} \geqslant 4$,

Equality holds if and only if $e^{x-a}=4e^{a-x}$, that is, when $x=a+\ln 2$,

Therefore, $f(x)-g(x) \geqslant 3$, and equality holds if and only if equality holds simultaneously,

Since there exists a real number $x_0$ such that $f(x_0)-g(x_0)=3$,

Thus, $a+\ln 2=-1$,

That is, $a=-1-\ln 2$.

Therefore, the answer is $\boxed{-1-\ln 2}$.