$1.4 \frac{3 a^{2}+2 a x-x^{2}}{(3 x+a)(a+x)}-2+10 \frac{a x-3 x^{2}}{a^{2}+9 x^{2}}$.

$1.5\left(\left(\frac{x}{y-x}\right)^{-2}-\frac{(x+y)^{2}-4 x y}{x^{2}-x y}\right)^{2} \frac{x^{4}}{x^{2} y^{2}-y^{4}}$.
1.4 According to formula (1.8), we have $a^{2}-9 x^{2}=(a-3 x)(a+3 x)$. Simplify the fraction $\frac{a x-3 x^{2}}{(a-3 x)(a+3 x)}$ by canceling $a-3 x$, assuming $a \neq 3 x$; we get $\frac{x}{a+3 x}$. Next, using formula (1.15), factor the numerator of the first fraction:

$$
\begin{aligned}
& 3 a^{2}+2 a x-x^{2}=-\left(x^{2}-2 a x-3 a^{2}\right)=-(x-3 a)(x+a)= \\
& =(3 a-x)(x+a)
\end{aligned}
$$

Now, cancel the first fraction by $a+x$, where $a \neq -x$, and we find

$$
\frac{3 a-x}{3 x+a}-2+\frac{10 x}{a+3 x}=\frac{3 a-x-2(3 x+a)+10 x}{a+3 x}=\frac{a+3 x}{a+3 x}=1
$$

Answer: 1.