Given that \(\tan \alpha = 3\) and \(\frac{\pi}{4} < \alpha \leq \frac{3\pi}{2}\), we need to determine \(\sin \alpha\), \(\cos \alpha\), and \(\cot \alpha\).

1. **Find \(\sin \alpha\) and \(\cos \alpha\)**:
- From \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = 3\), we can express this as:
\[
\sin \alpha = 3 \cos \alpha
\]
- Using the Pythagorean identity \(\sin^2 \alpha + \cos^2 \alpha = 1\), we substitute \(\sin \alpha\):
\[
(3 \cos \alpha)^2 + \cos^2 \alpha = 1
\]
\[
9 \cos^2 \alpha + \cos^2 \alpha = 1
\]
\[
10 \cos^2 \alpha = 1
\]
\[
\cos^2 \alpha = \frac{1}{10}
\]
\[
\cos \alpha = -\sqrt{\frac{1}{10}} = -\frac{1}{\sqrt{10}}
\]
- Since \(\alpha\) is in the third quadrant, \(\cos \alpha\) is negative.

2. **Find \(\sin \alpha\)**:
- Substitute \(\cos \alpha\) back into the equation for \(\sin \alpha\):
\[
\sin \alpha = 3 \left(-\frac{1}{\sqrt{10}}\right) = -\frac{3}{\sqrt{10}}
\]

3. **Calculate \(\cot \alpha\)**:
- Using the definition of cotangent, which is the reciprocal of tangent:
\[
\cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{3}
\]

In summary, we have:
- \(\sin \alpha = -\frac{3}{\sqrt{10}}\)
- \(\cos \alpha = -\frac{1}{\sqrt{10}}\)
- \(\cot \alpha = \frac{1}{3}\)