Find the first three terms in ascending powers of y in the expansion of (1-y)^(1)/(2).

We use the binomial series expansion for $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$.
In this case, $x=-y$ and $n=\frac{1}{2}$.

The first term is:

$$1$$

The second term is $nx$:

$$\left(\frac{1}{2}\right)(-y) = -\frac{1}{2}y$$

The third term is $\frac{n(n-1)}{2!}x^2$:

$$\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}(-y)^2$$ $$= \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2 \times 1} (y^2)$$ $$= \frac{-\frac{1}{4}}{2} y^2$$ $$= -\frac{1}{8}y^2$$

The first three terms in ascending powers of $y$ are $\boxed{1 - \frac{1}{2}y - \frac{1}{8}y^2}$.

First, we expand $\left(2x-\frac{1}{x}\right)^3$ using the binomial expansion formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.
Let $a=2x$ and $b=-\frac{1}{x} = -x^{-1}$.

$$\left(2x-\frac{1}{x}\right)^3 = (2x)^3 + 3(2x)^2\left(-x^{-1}\right) + 3(2x)\left(-x^{-1}\right)^2 + \left(-x^{-1}\right)^3$$ $$= 8x^3 + 3(4x^2)(-x^{-1}) + 3(2x)(x^{-2}) + (-x^{-3})$$ $$= 8x^3 - 12x^{2-1} + 6x^{1-2} - x^{-3}$$ $$= 8x^3 - 12x + 6x^{-1} - x^{-3}$$

Now, we need to find the coefficient of $x^{-1}$ in the expansion of $(x+1)\left(8x^3 - 12x + 6x^{-1} - x^{-3}\right)$.
We multiply each term in the first bracket by each term in the second bracket and identify terms that result in $x^{-1}$.

Terms from $x \times \left(8x^3 - 12x + 6x^{-1} - x^{-3}\right)$:

• $x \cdot (8x^3) = 8x^4$
• $x \cdot (-12x) = -12x^2$
• $x \cdot (6x^{-1}) = 6x^0 = 6$
• $x \cdot (-x^{-3}) = -x^{-2}$ None of these terms result in $x^{-1}$.

Terms from $1 \times \left(8x^3 - 12x + 6x^{-1} - x^{-3}\right)$:

• $1 \cdot (8x^3) = 8x^3$
• $1 \cdot (-12x) = -12x$
• $1 \cdot (6x^{-1}) = 6x^{-1}$ (This term has $x^{-1}$)
• $1 \cdot (-x^{-3}) = -x^{-3}$ The only term that results in $x^{-1}$ is $6x^{-1}$. Therefore, the coefficient of $x^{-1}$ is $\boxed{6}$.