已知 $\frac{\mathrm{d}w}{\mathrm{d}x} = (w - 1)(2 - w), \quad \text{for } 1 < w < 2. \quad \mathbf{(I)}$ 又已知 $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 - (3 + 2x)\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) + x^2 + 3x + 1 = 0, \quad\text{for }\; 1 < \frac{\mathrm{d}y}{\mathrm{d}x} - x < 2. \quad (\mathrm{II})$

1. Separate the variables $w$ and $x$.

$\int \frac{1}{(w - 1)(2 - w)} \, \mathrm{d}w = \int 1 \, \mathrm{d}x$

2. Express the integrand in partial fractions.

$\frac{1}{(w - 1)(2 - w)} \equiv \frac{P}{w - 1} + \frac{Q}{2 - w}$
$1 = P(2 - w) + Q(w - 1)$
Substituting $w = 1$ gives $P = 1$.
Substituting $w = 2$ gives $Q = 1$.
$\int \left( \frac{1}{w - 1} + \frac{1}{2 - w} \right) \, \mathrm{d}w = \int 1 \, \mathrm{d}x$

3. Integrate both sides and combine the logarithmic terms.

$\ln(w - 1) - \ln(2 - w) = x + C$
(Note: Since $1 < w < 2$, $w - 1 > 0$ and $2 - w > 0$, so absolute value signs are not required.)
$\ln\left( \frac{w - 1}{2 - w} \right) = x + C$

4. Exponentiate both sides to remove the logarithm and define the constant $A$.

$\frac{w - 1}{2 - w} = \mathrm{e}^{x+C}$
$\frac{w - 1}{2 - w} = \mathrm{e}^C \mathrm{e}^x$
Let $A = \mathrm{e}^C$. Since $\mathrm{e}^C > 0$ for any real number $C$, $A$ is a positive constant.
$\frac{w - 1}{2 - w} = A\mathrm{e}^x$

5. Make $w$ the subject of the formula and rearrange it into the required form.

$w - 1 = A\mathrm{e}^x(2 - w)$
$w - 1 = 2A\mathrm{e}^x - wA\mathrm{e}^x$
$w + wA\mathrm{e}^x = 1 + 2A\mathrm{e}^x$
$w(1 + A\mathrm{e}^x) = 1 + 2A\mathrm{e}^x$
$w = \frac{1 + 2A\mathrm{e}^x}{1 + A\mathrm{e}^x}$
$w = \frac{1 + A\mathrm{e}^x + A\mathrm{e}^x}{1 + A\mathrm{e}^x}$
$w = 1 + \frac{A\mathrm{e}^x}{1 + A\mathrm{e}^x} \quad \text{(shown)}$

1. Find the horizontal asymptotes by evaluating the limits of $w$ as $x \to \pm\infty$.

As $x \to -\infty$, $\mathrm{e}^x \to 0$, so:
$w \to 1 + \frac{0}{1+0} = 1$

As $x \to \infty$, we can divide the numerator and denominator of the fraction by $\mathrm{e}^x$:
$w = 1 + \frac{A}{\mathrm{e}^{-x} + A}$
Since $\mathrm{e}^{-x} \to 0$ as $x \to \infty$, we have:
$w \to 1 + \frac{A}{0 + A} = 1 + 1 = 2$

Therefore, the horizontal asymptotes are $w = 1$ and $w = 2$.

2. Find the coordinates of the axes intercepts.

To find the $w$-intercept, substitute $x = 0$:
$w = 1 + \frac{A\mathrm{e}^0}{1+A\mathrm{e}^0} = 1 + \frac{A}{1+A} = \frac{1+A+A}{1+A} = \frac{1+2A}{1+A}$
The curve meets the $w$-axis at $\left(0, \frac{1+2A}{1+A}\right)$.

To find the $x$-intercept, we set $w = 0$. However, the question states that $1 < w < 2$ for all real $x$ (or alternatively, setting $w=0$ yields $A\mathrm{e}^x = -\frac{1}{2}$, which has no real solutions since $A>0$ and $\mathrm{e}^x>0$).
Thus, the curve does not meet the $x$-axis.

3. Describe the key features for sketching the graph.

The graph of $w$ against $x$ is an S-shaped (logistic) curve.
It is strictly increasing from the lower asymptote $w = 1$ (as $x \to -\infty$) to the upper asymptote $w = 2$ (as $x \to \infty$).
It crosses the $w$-axis exactly once at the point $\left(0, \frac{1+2A}{1+A}\right)$, which lies between $w = 1$ and $w = 2$.

1. Differentiate the substitution equation with respect to $x$ to find the second derivative.

Given $\frac{\mathrm{d}y}{\mathrm{d}x} = w + x$, differentiating with respect to $x$ gives:
$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}w}{\mathrm{d}x} + 1$

2. Substitute the expressions for the first and second derivatives into equation (II).

Substituting into (II):
$\left(\frac{\mathrm{d}w}{\mathrm{d}x} + 1\right) + (w + x)^2 - (3 + 2x)(w + x) + x^2 + 3x + 1 = 0$

3. Expand and simplify the equation to show it reduces to equation (I), verifying the condition for $w$.

Expanding the terms:
$\frac{\mathrm{d}w}{\mathrm{d}x} + 1 + w^2 + 2wx + x^2 - 3w - 3x - 2wx - 2x^2 + x^2 + 3x + 1 = 0$
Collecting like terms ($x^2$, $x$, and $wx$ terms all cancel out):
$\frac{\mathrm{d}w}{\mathrm{d}x} + w^2 - 3w + 2 = 0$
$\frac{\mathrm{d}w}{\mathrm{d}x} = -w^2 + 3w - 2$
$\frac{\mathrm{d}w}{\mathrm{d}x} = -(w^2 - 3w + 2) = -(w - 1)(w - 2)$
$\frac{\mathrm{d}w}{\mathrm{d}x} = (w - 1)(2 - w)$
Additionally, the constraint $1 < \frac{\mathrm{d}y}{\mathrm{d}x} - x < 2$ becomes $1 < w < 2$.
This perfectly matches equation (I).

4. Set up a differential equation for $y$ using the solution for $w$ from part (a).

From part (a), we have $w = 1 + \frac{A\mathrm{e}^x}{1 + A\mathrm{e}^x}$.
Substituting this back into $\frac{\mathrm{d}y}{\mathrm{d}x} = w + x$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + \frac{A\mathrm{e}^x}{1 + A\mathrm{e}^x} + x$

5. Integrate with respect to $x$ to find $y$ in terms of $x$, recognizing the standard integrand form $\frac{\mathrm{f}'(x)}{\mathrm{f}(x)}$.

Integrating with respect to $x$:
$y = \int \left( 1 + x + \frac{A\mathrm{e}^x}{1 + A\mathrm{e}^x} \right) \mathrm{d}x$
$y = x + \frac{1}{2}x^2 + \ln(1 + A\mathrm{e}^x) + C$
where $C$ is an arbitrary constant. (Note: Since $A > 0$ and $\mathrm{e}^x > 0$, $1 + A\mathrm{e}^x > 1 > 0$, so no modulus is needed for the natural logarithm).