(a) Start with

\[ y^2=k\frac{x-2}{x+2}. \]

Differentiate implicitly:

\[ 2y\frac{dy}{dx} = k\frac{(x+2)-(x-2)}{(x+2)^2}. \]

So

\[ 2y\frac{dy}{dx}=\frac{4k}{(x+2)^2}. \]

From the original equation,

\[ k=\frac{y^2(x+2)}{x-2}. \]

Substitute this into the derivative equation:

\[ 2y\frac{dy}{dx} = \frac{4}{(x+2)^2}\cdot\frac{y^2(x+2)}{x-2}. \]

\[ 2y\frac{dy}{dx}=\frac{4y^2}{(x+2)(x-2)}. \]

\[ 2y\frac{dy}{dx}=\frac{4y^2}{x^2-4}. \]

Therefore

\[ \frac{dy}{dx}=\frac{2y}{x^2-4}. \]

(b) When \(k=5\) and \(x=3\),

\[ y^2=5\frac{3-2}{3+2}=1. \]

So

\[ y=1\quad\text{or}\quad y=-1. \]

Using

\[ \frac{dy}{dx}=\frac{2y}{x^2-4}, \]

at \(x=3\):

\[ \frac{dy}{dx}=\frac{2y}{9-4}=\frac{2y}{5}. \]

So the two gradients are

\[ m_1=\frac25,\quad m_2=-\frac25. \]

The tangents make angles

\[ \tan^{-1}\left(\frac25\right) \]

and

\[ -\tan^{-1}\left(\frac25\right) \]

with the positive \(x\)-axis.

Therefore the angle between the tangents is

\[ 2\tan^{-1}\left(\frac25\right). \]

Answer: \(2\tan^{-1}\left(\frac25\right)\).