曲线 C 由参数方程
$x=\dfrac{t}{1+t},\quad y=\dfrac{t^{2}}{1+t}$，
其中 $t$ 取除 $-1$ 以外的所有实数值。

1. Find the derivative of $x$ with respect to $t$.

Given $x = \dfrac{t}{1+t}$, using the quotient rule:
$\frac{dx}{dt} = \frac{1(1+t) - t(1)}{(1+t)^2} = \frac{1}{(1+t)^2}$

2. Find the derivative of $y$ with respect to $t$.

Given $y = \dfrac{t^2}{1+t}$, using the quotient rule:
$\frac{dy}{dt} = \frac{2t(1+t) - t^2(1)}{(1+t)^2} = \frac{2t + 2t^2 - t^2}{(1+t)^2} = \frac{t^2 + 2t}{(1+t)^2}$

3. Apply the chain rule for parametric equations to find $\dfrac{dy}{dx}$.

Using the chain rule $\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$:
$\frac{dy}{dx} = \frac{\dfrac{t^2 + 2t}{(1+t)^2}}{\dfrac{1}{(1+t)^2}}$
$\frac{dy}{dx} = t^2 + 2t$

1. Find the derivative $\frac{dy}{dx}$ parametrically (from part (a) dependency).

Using the quotient rule:
$\frac{dx}{dt} = \frac{(1+t)(1) - t(1)}{(1+t)^2} = \frac{1}{(1+t)^2}$
$\frac{dy}{dt} = \frac{(1+t)(2t) - t^2(1)}{(1+t)^2} = \frac{t^2+2t}{(1+t)^2}$
Using the chain rule:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = t^2+2t = t(t+2)$

2. Evaluate the gradient of the tangent at the specified point.

At the point $\left(\frac{p}{1+p},\;\frac{p^2}{1+p}\right)$, the parameter is $t = p$.
Substituting $t = p$ into $\frac{dy}{dx}$, the gradient of the tangent $m$ is:
$m = p(p+2)$

3. Formulate the equation of the tangent and simplify it to the required form.

Using the point-gradient equation of a straight line, $y - y_1 = m(x - x_1)$:
$y - \frac{p^2}{1+p} = p(p+2) \left( x - \frac{p}{1+p} \right)$
Expanding and rearranging for $y$:
$y = p(p+2)x - \frac{p^2(p+2)}{1+p} + \frac{p^2}{1+p}$
$y = p(p+2)x - \frac{p^3+2p^2-p^2}{1+p}$
$y = p(p+2)x - \frac{p^3+p^2}{1+p}$
Factorising the numerator of the fraction:
$y = p(p+2)x - \frac{p^2(p+1)}{1+p}$
$y = p(p+2)x - p^2 \quad \text{(shown)}$

1. Substitute the point $(2,5)$ into the tangent equation to form an equation in terms of $p$.

Since the tangent passes through $(2,5)$, we substitute $x=2$ and $y=5$ into the given equation of the tangent $y = p(p+2)x - p^2$:
$5 = p(p+2)(2) - p^2$

2. Solve the quadratic equation to find the values of $p$.

$5 = 2p^2 + 4p - p^2$
$p^2 + 4p - 5 = 0$
$(p+5)(p-1) = 0$
$p = -5 \text{ or } p = 1$

3. Find the gradient of each tangent line.

The gradient of the tangent is given by the coefficient of $x$, which is $m = p(p+2)$.
For $p = -5$, $m_1 = -5(-5+2) = 15$.
For $p = 1$, $m_2 = 1(1+2) = 3$.

4. Calculate the acute angle between the two tangents.

Let $\theta$ be the acute angle between the two lines.
Using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$:
$\tan \theta = \left| \frac{15 - 3}{1 + (15)(3)} \right|$
$\tan \theta = \frac{12}{46} = \frac{6}{23}$
$\theta = \tan^{-1}\left(\frac{6}{23}\right) \approx 14.6^\circ$