Step 1: Find the binomial expansion of $(x-2y)^3$. We use the binomial expansion formula $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$. For $n=3$, the expansion is $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. Substitute $a=x$ and $b=-2y$: $$ (x-2y)^3 = (x)^3 + 3(x)^2(-2y) + 3(x)(-2y)^2 + (-2y)^3 $$ $$ (x-2y)^3 = x^3 + 3x^2(-2y) + 3x(4y^2) + (-8y^3) $$ $$ (x-2y)^3 = x^3 - 6x^2y + 12xy^2 - 8y^3 $$ The binomial expansion is $\boxed{x^3 - 6x^2y + 12xy^2 - 8y^3}$. Step 2: Find the term independent of $x$ in the expansion of $\left(x - \frac{3}{x}\right)^8$. The general term in the binomial expansion of $(a+b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. Here, $a=x$, $b=-\frac{3}{x} = -3x^{-1}$, and $n=8$. Substitute these values into the general term formula: $$ T_{r+1} = \binom{8}{r} (x)^{8-r} (-3x^{-1})^r $$ $$ T_{r+1} = \binom{8}{r} x^{8-r} (-3)^r (x^{-1})^r $$ $$ T_{r+1} = \binom{8}{r} (-3)^r x^{8-r-r} $$ $$ T_{r+1} = \binom{8}{r} (-3)^r x^{8-2r} $$ For the term to be independent of $x$, the power of $x$ must be 0. $$ 8-2r = 0 $$ $$ 2r = 8 $$ $$ r = 4 $$ Now, substitute $r=4$ back into the general term: $$ T_{4+1} = T_5 = \binom{8}{4} (-3)^4 x^{8-2(4)} $$ $$ T_5 = \binom{8}{4} (-3)^4 x^0 $$ Calculate $\binom{8}{4}$: $$ \binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 $$ Calculate $(-3)^4$: $$ (-3)^4 = 81 $$ Now, multiply these values: $$ T_5 = 70 \times 81 = 5670 $$ The term independent of $x$ is $\boxed{5670}$. Step 3: Find the first three terms in ascending powers of $y$ in the expansion of $(1-y)^{\frac{1}{2}}$. We use the binomial series expansion for $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$. Here, $x=-y$ and $n=\frac{1}{2}$. First term: $1$ Second term: $nx = \left(\frac{1}{2}\right)(-y) = -\frac{1}{2}y$ Third term: $\frac{n(n-1)}{2!}x^2 = \frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}(-y)^2$ $$ = \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2} (y^2) $$ $$ = \frac{-\frac{1}{4}}{2} y^2 $$ $$ = -\frac{1}{8}y^2 $$ The first three terms are $\boxed{1 - \frac{1}{2}y - \frac{1}{8}y^2}$. Step 4: Find the coefficient of $x^{-1}$ in the expansion of $\left((x+1)\left(2x-\frac{1}{x}\right)^3\right)$. First, expand $\left(2x-\frac{1}{x}\right)^3$ using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. Let $a=2x$ and $b=-\frac{1}{x}$: $$ \left(2x-\frac{1}{x}\right)^3 = (2x)^3 + 3(2x)^2\left(-\frac{1}{x}\right) +