[THOUGHT] Let me solve this problem step by step.

Part (1): Area enclosed by the curve For a polar curve r = f(θ), the area is given by:

Part (2): Slope of tangent line For a polar curve, the slope in Cartesian coordinates is:

Given r = 2 + cos(θ), we have dr/dθ = -sin(θ)

At θ = π/4, I need to compute this slope.

Let me calculate both parts using Python. [/THOUGHT] [PYTHON] import sympy as sp from sympy import pi, cos, sin, sqrt, simplify, integrate, Rational

theta = sp.Symbol('theta')

Part (1): Area calculation

r = 2 + cos(theta) r_squared = r**2

Expand r^2

r_squared_expanded = sp.expand(r_squared) print("r^2 =", r_squared_expanded)

Area = (1/2) * integral from 0 to 2*pi of r^2 dtheta

area_integral = integrate(r_squared, (theta, 0, 2*pi)) area = Rational(1, 2) * area_integral print("Area =", area) print("Area simplified =", simplify(area))

Part (2): Slope of tangent line at theta = pi/4

dy/dx = (dr/dtheta * sin(theta) + r * cos(theta)) / (dr/dtheta * cos(theta) - r * sin(theta))

dr_dtheta = sp.diff(r, theta) print("\ndr/dtheta =", dr_dtheta)

At theta = pi/4

theta_val = pi/4

r_val = r.subs(theta, theta_val) dr_dtheta_val = dr_dtheta.subs(theta, theta_val)

print("r at pi/4 =", r_val) print("dr/dtheta at pi/4 =", dr_dtheta_val)

Numerator: dr/dtheta * sin(theta) + r * cos(theta)

numerator = dr_dtheta_val * sin(theta_val) + r_val * cos(theta_val)

Denominator: dr/dtheta * cos(theta) - r * sin(theta)

denominator = dr_dtheta_val * cos(theta_val) - r_val * sin(theta_val)

print("\nNumerator =", numerator) print("Denominator =", denominator)

slope = numerator / denominator slope_simplified = simplify(slope) print("\nSlope =", slope_simplified)

Let's rationalize and simplify further

slope_rationalized = sp.nsimplify(slope_simplified, rational=False) print("Slope rationalized =", slope_rationalized)

Let's expand this more carefully

slope_expanded = sp.simplify(sp.expand(slope_simplified)) print("Slope expanded =", slope_expanded) [/PYTHON]

[THOUGHT] From the Python calculations:

Part (1): The area enclosed by the curve is $a = \frac{9\pi}{2}$

Part (2): The slope of the tangent line is $-2 + \sqrt{2}$, which can be written as $b + \sqrt{c}$ where $b = -2$ and $c = 2$.

Now I need to compute:

Wait, let me recalculate:

The answer is $-18$. [/THOUGHT]

$\boxed{-18}$