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., $T_{r-3+1}$.

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

Given expansion is $(1+x)^{2 n}$.

Now,

$$\begin{aligned} & \text { coefficient of } 2 \text { nd term }={ }^{2 n} \mathrm{C}_1 \\ & \text { Coefficient of 3rd term }={ }^{2 n} \mathrm{C}_2 \\ & \text { Coefficient of 4th term }={ }^{2 n} C_3 \end{aligned}$$

Given that, ${ }^{2 n} C_1,{ }^{2 n} C_2$ and ${ }^{2 n} C_3$ are in AP.

$$\begin{aligned} & \text { Then, } & 2^{2{ }^{2 n} C_2} & ={ }^{2 n} C_1+{ }^{2 n} C_3 \\ \Rightarrow & & 2\left[\frac{2 n(2 n-1)(2 n-2)!}{2 \times 1 \times(2 n-2)!}\right] & =\frac{2 n(2 n-1)!}{(2 n-1)!}+\frac{2 n(2 n-1)(2 n-2)(2 n-3)!}{3!(2 n-3)!} \\ \Rightarrow & & n(2 n-1) & =n+\frac{n(2 n-1)(2 n-2)}{6} \\ \Rightarrow & & n(12 n-6) & =n\left(6+4 n^2-4 n-2 n+2\right) \\ \Rightarrow & & 12 n-6 & =\left(4 n^2-6 n+8\right) \\ \Rightarrow & & 6(2 n-1) & =2\left(2 n^2-3 n+4\right) \\ \Rightarrow & & 3(2 n-1) & =2 n^2-3 n+4 \\ \Rightarrow & & 2 n^2-3 n+4-6 n+3 & =0 \\ \Rightarrow & & 2 n^2-9 n+7 & =0 \end{aligned}$$

Given, expansion

$$\begin{aligned} & =\left(1+x+x^2+x^3\right)^{11}=\left[(1+x)+x^2(1+x)\right]^{11} \\ & =\left[(1+x)\left(1+x^2\right)\right]^{11}=(1+x)^{11} \cdot\left(1+x^2\right)^{11} \end{aligned}$$

Now, above expansion becomes

$$\begin{aligned} & =\left({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+{ }^{11} C_3 x^3+{ }^{11} C_4 x^4+\ldots\right)\left({ }^{11} C_0+{ }^{11} C_1 x^2+{ }^{11} C_2 x^4+\ldots\right) \\ & =\left(1+11 x+55 x^2+165 x^3+330 x^4+\ldots\right)\left(1+11 x^2+55 x^4+\ldots\right) \end{aligned}$$

$\therefore$ Coefficient of $x^4=55+605+330=990$

Given expansion is $\left(\frac{p}{2}+2\right)^8$.

Here, $n=8\quad$ [even]

Since, this Binomial expansion has only one middle term i.e., $\left(\frac{8}{2}+1\right)$ th $=5$ th term

$$\begin{array}{l} & T_5 & =T_{4+1}={ }^8 C_4\left(\frac{p}{2}\right)^{8-4} \cdot 2^4 \\ \Rightarrow \quad & 1120 & ={ }^8 C_4 p^4 \cdot 2^{-4} 2^4 \\ \Rightarrow \quad & 1120 & =\frac{8 \times 7 \times 6 \times 5 \times 4!}{4!\times 4 \times 3 \times 2 \times 1} p^4 \end{array}$$

$$\begin{aligned} & \Rightarrow \quad 1120=7 \times 2 \times 5 \times p^4 \\ & \Rightarrow \quad p^4=\frac{1120}{70}=16 \Rightarrow p^4=2^4 \\ & \Rightarrow \quad p^2=4 \Rightarrow p= \pm 2 \end{aligned}$$