The gradient function of a curve $y = f(x)$ is given by $\frac{36}{(2x+1)^2}$ for $x > -\frac{1}{2}$. The curve passes through the point $\left(\frac{1}{2}, 2\right)$.

1. Integrate the gradient function to find the general equation of the curve

$y = \int \frac{36}{(2x+1)^2} \, dx$
$y = \int 36(2x+1)^{-2} \, dx$
$y = \frac{36(2x+1)^{-1}}{(-1)(2)} + c$
$y = -\frac{18}{2x+1} + c$

2. Substitute the given point to find the constant of integration

Substitute $x = \frac{1}{2}$ and $y = 2$:
$2 = -\frac{18}{2(\frac{1}{2})+1} + c$
$2 = -\frac{18}{2} + c$
$2 = -9 + c$
$c = 11$

3. State the final equation of the curve

$y = -\frac{18}{2x+1} + 11$

1. Find the gradient of the tangent at $x = 1$.

At $x = 1$,
$\frac{dy}{dx} = \frac{36}{(2(1)+1)^2}$
$\frac{dy}{dx} = \frac{36}{3^2} = \frac{36}{9} = 4$

2. Calculate the angle the tangent makes with the $x$-axis.

Let $\theta$ be the angle the tangent makes with the $x$-axis.
$\tan \theta = 4$
$\theta = \tan^{-1}(4)$
$\theta \approx 75.964^\circ$
$\theta = 76.0^\circ \text{ (to 1 d.p.)}$