Sin 13пи/2-cos 3пи/2+cos12пи-tg*+ctg(-3пи/2)

To simplify \( \sin \frac{13\pi}{2} - \cos \frac{3\pi}{2} + \cos 12\pi - \tan + \cot (-\frac{3\pi}{2}) \), we will evaluate each term step by step:

1. Evaluate \( \sin \frac{13\pi}{2} \): \[ \frac{13\pi}{2} = 6\pi + \frac{\pi}{2} \] Since sine has a period of \( 2\pi \), we can reduce this: \[ \sin \frac{13\pi}{2} = \sin \left(\frac{\pi}{2}\right) = 1 \]

2. Evaluate \( -\cos \frac{3\pi}{2} \): \[ \cos \frac{3\pi}{2} = 0, \quad \text{thus} \quad -\cos \frac{3\pi}{2} = -0 = 0 \]

3. Evaluate \( \cos 12\pi \): \[ \cos 12\pi = \cos(6 \cdot 2\pi) = 1 \]

4. Evaluate \( \tan \): It seems like this term is incomplete or missing arguments. Assuming it’s a typo or an incomplete function, we can consider it as \( 0 \) for simplification if needed.

5. Evaluate \( \cot (-\frac{3\pi}{2}) \): The cotangent function is defined as \( \cot x = \frac{\cos x}{\sin x} \). \[ \cot(-\frac{3\pi}{2}) = \frac{\cos(-\frac{3\pi}{2})}{\sin(-\frac{3\pi}{2})} = \frac{0}{1} = 0 \]

Now, let's combine everything: \[ \sin \frac{13\pi}{2} - \cos \frac{3\pi}{2} + \cos 12\pi - \tan + \cot (-\frac{3\pi}{2}) = 1 - 0 + 1 - 0 + 0 \]

So the final answer is: \[ 1 + 1 = 2 \]

Thus, the result of the expression is: \[ \boxed{2} \]