Given expansion is $\left(\sqrt{x}-\frac{k}{x^2}\right)^{10}$.

Let $T_{t+1}$ is the general term.

Then,

$$\begin{aligned} T_{r+1} & ={ }^{10} C_r(\sqrt{x})^{10-r}\left(\frac{-k}{x^2}\right)^r \\ & ={ }^{10} C_r(x)^{\frac{1}{2}(10-r)}(-k)^r \cdot x^{-2 r} \\ & ={ }^{10} C_r x^{5-\frac{r}{2}}(-k)^r \cdot x^{-2 r} \\ & ={ }^{10} C_r x^{5-\frac{r}{2}-2 r}(-k)^r \\ & ={ }^{10} C_r x^{\frac{10-5 r}{2}}(-k)^r \end{aligned}$$

For free from $x$,

$$\begin{aligned} \frac{10-5 r}{2} & =0 \\ 10-5 r & =0 \Rightarrow r=2 \end{aligned}$$

Since, $T_{2+1}=T_3$ is free from $x$.

$$\begin{array}{ll} \therefore & T_{2+1}={ }^{10} C_2(-k)^2=405 \\ \Rightarrow & \frac{10 \times 9 \times 8!}{2!\times 8!}(-k)^2=405 \\ \Rightarrow & 45 k^2=405 \Rightarrow k^2=\frac{405}{45}=9 \\ \therefore & k= \pm 3 \end{array}$$

Given,

$$\begin{aligned} \text { expansion } & =\left(1-3 x+7 x^2\right)(1-x)^{16} \\ & =\left(1-3 x+7 x^2\right)\left({ }^{16} C_0 1^{16}-{ }^{16} C_1 1^{15} x^1+{ }^{16} C_2 1^{14} x^2+\ldots+{ }^{16} C_{16} x^{16}\right) \\ & =\left(1-3 x+7 x^2\right)\left(1-16 x+120 x^2+\ldots\right) \end{aligned}$$

$\therefore \quad$ Coefficient of $x=-3-16=-19$

Given expansion is $\left(3 x-\frac{2}{x^2}\right)^{15}$.

Let $T_{r+1}$ is the general term.

$$\begin{aligned} \therefore \quad T_{r+1} & ={ }^{15} C_r(3 x)^{15-r}\left(\frac{-2}{x^2}\right)^r={ }^{15} C_r(3 x)^{15-r}(-2)^r x^{-2 r} \\ & ={ }^{15} C_r 3^{15-r} x^{15-3 r}(-2)^r \end{aligned}$$

For independent of $x, \quad 15-3 r=0 \Rightarrow r=5$

Since, $T_{5+1}=T_6$ is independent of $x$.

$$\begin{aligned} \therefore \quad T_{5+1} & ={ }^{15} C_5 3^{15-5}(-2)^5 \\ & =-\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10!}{5 \times 4 \times 3 \times 2 \times 1 \times 10!} \cdot 3^{10} \cdot 2^5 \\ & =-3003 \cdot 3^{10} \cdot 2^5 \end{aligned}$$

(i) Given expansion is $\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$.

Here, the power of Binomial i.e., $n=10$ is even.

Since, it has one middle term $\left(\frac{10}{2}+1\right)$ th term i.e., 6th term.

$$\begin{aligned} \therefore \quad T_6 & =T_{5+1}={ }^{10} C_5\left(\frac{x}{a}\right)^{10-5}\left(\frac{-a}{x}\right)^5 \\ & =-{ }^{10} C_5\left(\frac{x}{a}\right)^5\left(\frac{a}{x}\right)^5 \\ & =-\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5!\times 5 \times 4 \times 3 \times 2 \times 1}\left(\frac{x}{a}\right)^5\left(\frac{x}{a}\right)^{-5} \\ & =-9 \times 4 \times 7=-252 \end{aligned}$$

(ii) Given expansion is $\left(3 x-\frac{x^3}{6}\right)^9$.

Here, $n=9$ [odd]

Since, the Binomial expansion has two middle terms i.e., $\left(\frac{9+1}{2}\right)$ th and $\left(\frac{9+1}{2}+1\right)$ th i.e., 5 th term and 6 th term.

$$\begin{aligned} \therefore \quad T_5 & =T_{(4+1)}={ }^9 C_4(3 x)^{9-4}\left(-\frac{x^3}{6}\right)^4 \\ & =\frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 1 \times 5!} 3^5 x^5 x^{12} 6^{-4} \\ & =\frac{7 \times 6 \times 3 \times 3^1}{2^4} x^{17}=\frac{189}{8} x^{17} \end{aligned}$$

$$ \begin{aligned} & \therefore \quad T_6=T_{5+1}={ }^9 C_5(3 x)^{9-5}\left(-\frac{x^3}{6}\right)^5 \\ & =-\frac{9 \times 8 \times 7 \times 6 \times 5!}{5!\times 4 \times 3 \times 2 \times 1} \cdot 3^4 \cdot x^4 \cdot x^{15} \cdot 6^{-5} \\ & =\frac{-21 \times 6}{3 \times 2^5} x^{19}=\frac{-21}{16} x^{19} \end{aligned}$$

Given expansion is $\left(x-x^2\right)^{10}$.

Let the term $T_{r+1}$ is the general term.

$$\begin{aligned} \therefore \quad T_{r+1} & ={ }^{10} C_r x^{10-r}\left(-x^2\right)^r \\ & =(-1)^r \cdot{ }^{10} C_r \cdot x^{10-r} \cdot x^{2 r} \\ & =(-1)^{10} C_r x^{10+r} \end{aligned}$$

For the coefficient of $x^{15}$,

$$\begin{aligned} 10+r & =15 \Rightarrow r=5 \\ T_{5+1} & =(-1)^5{ }^{10} C_5 x^{15} \\ \therefore \quad \text { Coefficient of } x^{15} & =-1 \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5 \times 4 \times 3 \times 2 \times 1 \times 5!} \\ & =-3 \times 2 \times 7 \times 6=-252 \end{aligned}$$