1. Let $a \in \mathbf{R}_{+}$. If the function
$$
f(x)=\frac{a}{x-1}+\frac{1}{x-2}+\frac{1}{x-6}(3<x<5)
$$

attains its maximum value at $x=4$, find the value of $a$.
1. From the problem, for $30, x-2>0, x-6<0$. Clearly, $\frac{a}{x-1}+\frac{1}{x-2}+\frac{1}{x-6} \leqslant \frac{a}{3}$.
Eliminating the denominators, simplifying, and rearranging, we get
$$
\begin{array}{l}
a x^{3}-(12 a+6) x^{2}+(44 a+30) x-(48 a+24) \leqslant 0 \\
\Rightarrow(x-4)\left(a x^{2}-(8 a+6) x+(12 a+6)\right) \\
\quad \leqslant 0 .
\end{array}
$$

Since $x=4$ is the maximum value, the second factor on the left side of inequality (1) also has a root $x=4$.
Solving for $a$ gives $a=-\frac{9}{2}$.
Substituting $a$ into inequality (1) for verification, we get
$$
(x-4)^{2}\left(-\frac{9}{2} x+12\right) \leqslant 0 \text {, }
$$

which satisfies $3<x<5, x=4$ being the maximum value.