1. Let $f(x)=\mathrm{e}^{2 x}-1, g(x)=\ln (x+1)$. Then the solution set of the inequality $f(g(x))-g(f(x)) \leqslant 1$ is
$$
-1 .(-1,1] \text {. }
$$

Notice,
$$
f(g(x))=x^{2}+2 x, g(f(x))=2 x \text {. }
$$

Therefore, $f(g(x))-g(f(x))=x^{2}$.
The domain is defined as $(-1,+\infty)$, thus, the solution set of the inequality is $(-1,1]$.