If $x$ is so small that all terms containing $x^2$ and higher powers of $x$ can be neglected, then the approximate value of $\frac{\left(1+\frac{2 x}{3}\right)^{-4}(4+5 x)^{1 / 2}}{(9+x)^{3 / 2}}$, when $x=\frac{6}{371}$, is

The sum of the coefficients of $x^{-3 / 2}$ and $x^3$ in the expansion of $\sqrt{3+x}+\sqrt{5+x}$ when $3 < x< 5$, is

$p, q$ are two prime numbers. For $n=p q$, if the expansion $\left(\sqrt[4]{x^{-5}}+2 \sqrt[5]{x^5}\right)^n$ contains non-zero coefficient of $x^{-n}$ and $x^0$, then the least value of such $n$ is

The binomial expansion $(7+3 x)^{-2 / 5}$ is valid for all $x$ in the interval $\left(\frac{-7}{3}, \frac{7}{3}\right)$ and if the 4 th term of its expansion is $k x^3$, then $\left(7^{12 / 5} k\right)=$