(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 $x^{15}$,

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

Given expansion is $\left(x^4-\frac{1}{x^3}\right)^{15}$.

Let the term $T_{r+1}$ contains the coefficient of $\frac{1}{x^{17}}$ i.e., $x^{-17}$.

$$\begin{aligned} \therefore \quad T_{r+1} & ={ }^{15} C_r\left(x^4\right)^{15-r}\left(-\frac{1}{x^3}\right)^r \\ & ={ }^{15} C_r x^{60-4 r}(-1)^r x^{-3 r} \\ & ={ }^{15} C_r x^{60-7 r}(-1)^r \end{aligned}$$

For the coefficient $x^{-17}$,

$$\begin{aligned} & 60-7 r=-17 \\ & \begin{array}{lrl} \Rightarrow & 7 r & =77 \Rightarrow r=11 \\ \Rightarrow & T_{11+1} & ={ }^{15} C_{11} x^{60-77}(-1)^{11} \end{array} \\ & \therefore \quad \text { Coefficient of } x^{-17}=\frac{-15 \times 14 \times 13 \times 12 \times 11!}{11!\times 4 \times 3 \times 2 \times 1} \\ & =-15 \times 7 \times 13=-1365 \end{aligned}$$

Given expansion is $\left(y^{1 / 2}+x^{1 / 3}\right)^n$.

The sixth term of this expansion is

$$T_6=T_{5+1}={ }^n C_5\left(y^{1 / 2}\right)^{n-5}\left(x^{1 / 3}\right)^5\quad \text{... (i)}$$

Now, given that the Binomial coefficient of the third term from the end is 45 .

We know that, Binomial coefficient of third term from the end = Binomial coefficient of third term from the begining $={ }^n \mathrm{C}_2$

$$\begin{aligned} \because \quad & { }^n C_2 =45 \\ \Rightarrow \quad & \frac{n(n-1)(n-2)!}{2!(n-2)!} =45 \\ \Rightarrow \quad & n(n-1) =90 \\ \Rightarrow \quad & n^2-n-90 =0 \\ \Rightarrow \quad & n^2-10 n+9 n-90 =0 \\ \Rightarrow \quad & n(n-10)+9(n-10) =0 \\ \Rightarrow \quad & (n-10)(n+9) =0 \\ \Rightarrow \quad & (n+9) =0 \text { or }(n-10)=0 \\ \therefore \quad & n =10\quad [\because n \neq-9] \end{aligned}$$

From Eq. (i),

$$T_6={ }^{10} C_5 y^{5 / 2} x^{5 / 3}=252 y^{5 / 2} \cdot x^{5 / 3}$$

Given expansion is $(1+x)^{18}$.

Now, $(2 r+4)$ th term i.e., $T_{2 r+3+1}$.

$$\begin{aligned} \therefore \quad T_{2 r+3+1} & ={ }^{18} C_{2 r+3}(1)^{18-2 r-3}(x)^{2 r+3} \\ & ={ }^{18} C_{2 r+3} x^{2 r+3} \end{aligned}$$

Now, $(r-2)$ th term i.e., $T_{r-3+1}$.

$$\begin{array}{l} \therefore & T_{r-3+1} ={ }^{18} C_{r-3} x^{r-3} \\ \text { As, } & { }^{18} C_{2 r+3} ={ }^{18} C_{r-3} \quad \left[\because{ }^n C_x={ }^n C_y \Rightarrow x+y=n\right]\\ \Rightarrow & 2 r+3+r-3 =18 \\ \Rightarrow & 3 r =18 \\ \therefore & r =6 \end{array}$$