(a) Differentiate implicitly, applying the quotient rule on the right:

\[2y\frac{\mathrm{d}y}{\mathrm{d}x}=k\frac{(x+2)-(x-2)}{(x+2)^2}=\frac{4k}{(x+2)^2}.\]

From the given equation, \(k=y^2(x+2)/(x-2)\). Substitute this expression to eliminate \(k\):

\[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2k}{y(x+2)^2}=\frac{2y}{(x-2)(x+2)}=\frac{2y}{x^2-4},\]

as required.

(b) At \(x=3\), \(y^2=5(1/5)=1\), so \(y=\pm1\). The tangent gradients are therefore \(m_1=2/5\) and \(m_2=-2/5\).

For the acute angle \(\theta\) between the two lines,

\[\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|=\frac{4/5}{1-4/25}=\frac{20}{21}.\]

Hence \(\boxed{\theta=\tan^{-1}(20/21)}\). This has the requested form with \(a=20,b=1,c=21\).