Let's solve the two parts of the problem shown in the image.

Part (i)

We need to find the first three terms in the expansion of $(1 - 2x)^8$ in ascending powers of $x$.

To expand $(1 - 2x)^8$, we use the Binomial Theorem:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

Here, $a = 1$, $b = -2x$, and $n = 8$. The first three terms of the expansion will be for $k = 0$, $k = 1$, and $k = 2$.

1. Term for $k = 0$: $\binom{8}{0} \cdot 1^{8-0} \cdot (-2x)^0 = 1 \cdot 1 \cdot 1 = 1$

2. Term for $k = 1$: $\binom{8}{1} \cdot 1^{8-1} \cdot (-2x)^1 = 8 \cdot 1 \cdot (-2x) = -16x$

3. Term for $k = 2$: $\binom{8}{2} \cdot 1^{8-2} \cdot (-2x)^2 = \frac{8 \times 7}{2 \times 1} \cdot 1 \cdot 4x^2 = 28 \cdot 4x^2 = 112x^2$

Thus, the first three terms in the expansion of $(1 - 2x)^8$ are: $1 - 16x + 112x^2$

Part (ii)

Now, we need to find the coefficient of $x^2$ in the expansion of the expression:

$\frac{7 + 3x}{\sqrt{1 - 2x}}$

First, express $\sqrt{1 - 2x}$ as a binomial series. Using the Binomial Theorem for negative exponents:

$(1 - 2x)^{-\frac{1}{2}} = \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k} (-2x)^k$

The first few terms of the expansion are: $(1 - 2x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-2x) + \frac{-\frac{1}{2} \cdot -\frac{3}{2}}{2!}(-2x)^2 + \dots$ $= 1 + x + \frac{3x^2}{2} + \dots$

Now, consider the product: $\frac{7 + 3x}{\sqrt{1 - 2x}} = (7 + 3x)(1 + x + \frac{3x^2}{2} + \dots)$

Expanding and collecting the terms up to $x^2$: