1. For the real number $x$, the functions are
$$
f(x)=\sqrt{3 x^{2}+7}, g(x)=x^{2}+\frac{16}{x^{2}+1}-1,
$$

then the minimum value of the function $g(f(x))$ is . $\qquad$
$-, 1.8$.
From the problem, we have
$$
g(f(x))=3 x^{2}+7+\frac{16}{3 x^{2}+8}-1.
$$

Let $t=3 x^{2}+8(t \geqslant 8)$. Then
$$
h(t)=g(f(x))=t+\frac{16}{t}-2 \text{. }
$$

It is easy to see that $h(t)$ is a monotonically increasing function on the interval $[8,+\infty)$. Therefore, $h(t) \geqslant h(8)=8$.