It is given that $y = \sin\left(e^x - 1\right)$.

1. Differentiate $y$ with respect to $x$ using the chain rule to find $\frac{dy}{dx}$.

$y = \sin\left(e^x - 1\right)$
$\frac{dy}{dx} = \cos\left(e^x - 1\right) \cdot \frac{d}{dx}\left(e^x - 1\right)$
$\frac{dy}{dx} = e^x \cos\left(e^x - 1\right)$

2. Differentiate $\frac{dy}{dx}$ with respect to $x$ using the product rule to find $\frac{d^2 y}{dx^2}$.

$\frac{d^2 y}{dx^2} = \frac{d}{dx}\left(e^x\right) \cos\left(e^x - 1\right) + e^x \frac{d}{dx}\left(\cos\left(e^x - 1\right)\right)$
$\frac{d^2 y}{dx^2} = e^x \cos\left(e^x - 1\right) + e^x \left(-\sin\left(e^x - 1\right) \cdot e^x\right)$
$\frac{d^2 y}{dx^2} = e^x \cos\left(e^x - 1\right) - e^{2x} \sin\left(e^x - 1\right)$

3. Substitute $y$ and $\frac{dy}{dx}$ back into the second derivative expression to reach the required result.

Recognise that $\frac{dy}{dx} = e^x \cos\left(e^x - 1\right)$ and $y = \sin\left(e^x - 1\right)$. Substituting these into the second derivative:
$\frac{d^2 y}{dx^2} = \left(e^x \cos\left(e^x - 1\right)\right) - e^{2x} \left(\sin\left(e^x - 1\right)\right)$
$\frac{d^2 y}{dx^2} = \frac{dy}{dx} - e^{2x} y$
$\frac{d^2 y}{dx^2} = \frac{dy}{dx} - y e^{2x} \quad \text{(shown)}$

1. Evaluate $y$ and its first derivative at $x = 0$.

At $x = 0$, $y = \sin(e^0 - 1) = \sin(1 - 1) = \sin 0 = 0$.

Differentiating $y$ with respect to $x$ gives:
$\frac{dy}{dx} = e^x \cos(e^x - 1)$

At $x = 0$, $\frac{dy}{dx} = e^0 \cos(e^0 - 1) = 1 \cos 0 = 1$.

2. Use the identity from part (a) to evaluate the second derivative at $x = 0$.

From part (a), we are given:
$\frac{d^2 y}{dx^2} = \frac{dy}{dx} - ye^{2x}$

Substituting $x = 0$, $y = 0$, and $\frac{dy}{dx} = 1$:
$\frac{d^2 y}{dx^2} \Big|_{x=0} = 1 - (0)e^{2(0)} = 1$

3. Construct the Maclaurin series up to the term in $x^2$.

The Maclaurin series expansion is given by:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \dots$

Substituting the evaluated derivatives:
$y = 0 + 1(x) + \frac{1}{2!}(x^2)$
$y = x + \frac{1}{2}x^2$

1. Concisely compute the Maclaurin series of $y = \sin(e^x - 1)$ from part (b), as it is required to solve this part.

$y(0) = \sin(e^0 - 1) = 0$
$y' = e^x \cos(e^x - 1) \implies y'(0) = 1$
$y'' = y' - y e^{2x} \implies y''(0) = 1 - 0 = 1$
Thus, the Maclaurin series of $y$ is $y \approx 0 + 1x + \frac{1}{2!}x^2 = x + \frac{1}{2}x^2$.

2. Express $(1+y)^{\frac{1}{2}}$ using the standard binomial series expansion up to the $y^2$ term.

$\sqrt{1+y} = (1+y)^{\frac{1}{2}}$
Using the standard expansion for $(1+y)^n$ where $n = \frac{1}{2}$:
$(1+y)^{\frac{1}{2}} = 1 + \frac{1}{2}y + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}y^2 + \dots$
$= 1 + \frac{1}{2}y - \frac{1}{8}y^2 + \dots$

3. Substitute the series $y \approx x + \frac{1}{2}x^2$ into the binomial expansion and simplify, retaining terms up to $x^2$.

$\sqrt{1 + \sin(e^x - 1)} \approx 1 + \frac{1}{2}\left(x + \frac{1}{2}x^2\right) - \frac{1}{8}\left(x + \frac{1}{2}x^2\right)^2$
Dropping terms of order $x^3$ and higher:
$\approx 1 + \frac{1}{2}x + \frac{1}{4}x^2 - \frac{1}{8}(x^2)$
$= 1 + \frac{1}{2}x + \left(\frac{1}{4} - \frac{1}{8}\right)x^2$
$= 1 + \frac{1}{2}x + \frac{1}{8}x^2$