19. (1) Since $x, y \in \mathbf{R}^{+}$, we have $2=x+y \geqslant 2 \sqrt{x y}$, which means $x y \leqslant 1$.
$$\begin{array}{l} x^{3} y^{3}\left(x^{3}+y^{3}\right)=(x y)^{3}\left[(x+y)^{3}-3 x y(x+y)\right]= \\ (x y)^{3}(8-6 x y)=2(x y)^{3}(4-3 x y)= \\ 2[(x y)(x y)(x y)(4-3 x y)-] \leqslant 2\left[\frac{x y+x y+x y+(4-3 x y)}{4}\right]^{4}=2 \end{array}$$