Given: \(\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=12\). Let \(x=\sin\theta+\cos\theta\) and \(y=\sin\theta-\cos\theta\). Therefore, \(\frac{x}{y}=12\) implies \(x=12y\).

We know: \(x^2 = (\sin\theta+\cos\theta)^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 1 + 2\sin\theta\cos\theta\) and \(y^2 = (\sin\theta-\cos\theta)^2 = \sin^2\theta + \cos^2\theta - 2\sin\theta\cos\theta = 1 - 2\sin\theta\cos\theta\).

Using \(x=12y\), square both sides: \(x^2=(12y)^2=144y^2\). Therefore, \(\frac{x^2}{y^2}=\frac{144y^2}{y^2}=144\) so, \(\frac{1+2\sin\theta\cos\theta}{1-2\sin\theta\cos\theta}=144\).

Let \(z=2\sin\theta\cos\theta\). Then, \(\frac{1+z}{1-z}=144\). By cross-multiplying, we get \((1+z)=144(1-z)\), which simplifies to \(1+z=144-144z\). Thus, \(145z=143\), giving \(z=\frac{143}{145}\). So, \(2\sin\theta\cos\theta=\frac{143}{145}\), i.e., \(\sin 2\theta=\frac{143}{145}\).

\(\tan^2\theta=\frac{1-\cos 2\theta}{1+\cos 2\theta}=\frac{1-\frac{1}{145}}{1+\frac{1}{145}}\) simplifies as \(\frac{144/145}{146/145}=\frac{144}{146}=\frac{72}{73}\).

Similarly, \(\cot^2\theta=\frac{1+\cos 2\theta}{1-\cos 2\theta}=\frac{146/145}{144/145}=\frac{146}{144}=\frac{73}{72}\).

Now, evaluate the expression \(\frac{121\tan^2\theta-3}{169\cot^2\theta+1}\). Substitute \(\tan^2\theta\) and \(\cot^2\theta\):

Therefore, \(\frac{8493/73}{12409/72} =\frac{8493 \times 72}{12409 \times 73} = \frac{8493 \times 72}{12409 \times 73}=\frac{83}{61}\).

Thus, the answer is: 83/61.