Given that \(\cos A = \frac{2\sqrt{6}}{5}\) and \(A\) is in the interval \([90^\circ, 360^\circ]\), which implies \(A\) is in the fourth quadrant where \(\cos A > 0\) and \(\sin A < 0\). We use the Pythagorean identity to find \(\sin A\):

\[

\sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{2\sqrt{6}}{5}\right)^2

\]

Calculating \(\sin^2 A\) gives:

\[

\sin^2 A = 1 - \left(\frac{2\sqrt{6}}{5}\right)^2 = 1 - \frac{24}{25} = \frac{1}{25}

\]

Since \(A\) is in the fourth quadrant, \(\sin A\) is negative, thus:

\[

\sin A = -\sqrt{\frac{1}{25}} = -\frac{1}{5} = -0.2

\]

The tangent of \(A\) can be calculated as \(\tan A = \frac{\sin A}{\cos A}\), and thus:

\[

\tan A = \frac{-0.2}{\frac{2\sqrt{6}}{5}} = -\frac{1}{2\sqrt{6}} = -0.20412414523193162

\]

Finally, the value of \(5 \tan A \cdot \cos A\) is calculated as:

\[

5 \tan A \cdot \cos A = 5 \cdot (-0.20412414523193162) \cdot \frac{2\sqrt{6}}{5} = -1.0

\]

Therefore, \(5 \tan A \cdot \cos A = -1.0\).