[THOUGHT] I need to solve the problem step by step.

First, part (1): Compute the area enclosed by the curve $r = 2 + \cos\theta$ for $0 \le \theta \le 2\pi$.

In polar coordinates, the area enclosed by a curve $r(\theta)$ from $\theta = \alpha$ to $\theta = \beta$ is given by:

Here, $r(\theta) = 2 + \cos\theta$, and the curve is traced once as $\theta$ goes from 0 to $2\pi$. So:

Expand: $(2 + \cos\theta)^2 = 4 + 4\cos\theta + \cos^2\theta$.

Thus:

We can compute each integral separately.

Note: $\int_{0}^{2\pi} \cos\theta \, d\theta = 0$.

Also, $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$.

So:

The second integral is zero because $\cos 2\theta$ over a full period (actually two periods) integrates to zero. So:

Now:

So:

Thus:

So $a = \frac{9\pi}{2}$.

Now part (2): At the point $(r, \theta) = \left( \frac{4+\sqrt{2}}{2}, \frac{\pi}{4} \right)$, find the slope of the tangent line to the curve $C$ in the $xy$-plane.

We have $r = 2 + \cos\theta$. Check if the given point satisfies this: at $\theta = \pi/4$, $\cos(\pi/4) = \sqrt{2}/2$, so:

Yes, it matches.

We need the slope $\frac{dy}{dx}$ at that point. In polar coordinates:

Then:

We have:

Here, $r = 2 + \cos\theta$, so $\frac{dr}{d\theta} = -\sin\theta$.

Thus:

At $\theta = \pi/4$:

So:

Now compute:

Simplify:

Wait, let's compute carefully:

Divide by 4: $\frac{4\sqrt{2} + 2}{4} = \frac{2\sqrt{2} + 1}{2}$. So:

So $\frac{dy}{d\theta} = \sqrt{2}$.

Now $\frac{dx}{d\theta} = -\sin\theta \cos\theta - r \sin\theta$. At $\theta = \pi/4$:

So:

We already computed $r \cdot \frac{\sqrt{2}}{2} = \frac{2\sqrt{2} + 1}{2}$. Thus:

Therefore:

Simplify by multiplying numerator and denominator by $-1 + \sqrt{2}$:

So the slope is $-2 + \sqrt{2}$.

The problem asks: "Express the answer from part (2) in the form $b + \sqrt{c}$, where $b$ and $c$ are integers." Here, slope = $-2 + \sqrt{2}$, so $b = -2$, $c = 2$.

But wait: the problem says "let those integers be the definitions of $b$ and $c$." So $b = -2$, $c = 2$.

Now we need to compute $\frac{abc}{\pi}$, where $a = \frac{9\pi}{2}$, $b = -2$, $c = 2$.

Thus:

So the answer is -18.

Let me verify with Python to avoid mistakes.

[PYTHON] import sympy as sp

Part (1): area

theta = sp.symbols('theta') r = 2 + sp.cos(theta) area_expr = sp.Rational(1,2) * sp.integrate(r**2, (theta, 0, 2*sp.pi)) print("Area a =", area_expr) print("a =", area_expr.simplify())

Part (2): slope at given point

theta_val = sp.pi/4 r_val = (4 + sp.sqrt(2))/2 dr_dtheta = -sp.sin(theta_val) # derivative of r w.r.t theta

dy/dtheta and dx/dtheta

dy_dtheta = dr_dtheta * sp.sin(theta_val) + r_val * sp.cos(theta_val) dx_dtheta = dr_dtheta * sp.cos(theta_val) - r_val * sp.sin(theta_val)

slope = sp.simplify(dy_dtheta / dx_dtheta) print("Slope =", slope)

Express as b + sqrt(c)

slope_simplified = sp.simplify(slope) print("Simplified slope =", slope_simplified)

Identify b and c: slope = b + sqrt(c)

slope = -2 + sqrt(2)

b = -2 c = 2 print("b =", b, "c =", c)

Compute abc/pi

a = 9*sp.pi/2 abc_over_pi = (a * b * c) / sp.pi print("abc/pi =", abc_over_pi) print("abc/pi as integer =", abc_over_pi.simplify()) [/PYTHON]

[THOUGHT] The Python output confirms: area $a = \frac{9\pi}{2}$, slope = $-2 + \sqrt{2}$ so $b = -2$, $c = 2$, and $\frac{abc}{\pi} = -18$. The final answer is $-18$.

$\boxed{-18}$