II. (Full marks 15 points) $f(n)$ is a strictly increasing function defined on $\mathbf{N}$ and taking integer values (if for any $x_{1}, x_{2} \in A$, when $x_{1}<x_{2}$, we have $f\left(x_{1}\right)<f\left(x_{2}\right)$, then $f(x)$ is called a strictly increasing function on $A$). When $m, n$ are coprime, $f(m n)=f(m) f(n)$, and $f(19)=19$. Try to find $f(f(19) f(98))$.
II. H problem, $j(n)$ is defined on $\mathbf{N}$ with integer values and is a strictly increasing function.
$\because f(19)=f(1 \times 19)=f(1) \cdot f(19)$, but $f(19)=$ $19 \neq 0$,
$$
\therefore f(1)=1 \text {. }
$$

At this point, by $f(1)<f(2)<\cdots<f(19), f(1)=1$, $f(19)=19$, and $f(2), f(3), \cdots, f(18)$ are all natural numbers, so $f(n)=n(1 \leqslant n \leqslant 19, n \in \mathbf{N})$.

And $f(95)=f(5 \times 19)=f(5) f(19)=5 \times 19=$ 95,
$f(99)=f(9 \times 11)=f(9) f(11)=9 \times 11=99$.
Thus, $f(96)=96, f(97)=97, f(98)=98$.
Therefore, $f(f(19) f(98))=f(19 \times 98)$
$$
\begin{array}{l}
=f(19) f(98) \\
=19 \times 98=1862 .
\end{array}
$$