To find the maximum and minimum values (GTLN and GTNN) of the function \( y = \sqrt{2} \cos 2x + 4 \sin x \) on the interval \([0, \pi]\), we need to follow these steps:

1. Find the derivative of \( y \) with respect to \( x \).
2. Set the derivative equal to zero to find the critical points.
3. Evaluate the function at the critical points and at the endpoints of the interval.
4. Compare the values to determine the maximum and minimum.

Let's begin:

1. Derive \( y \) with respect to \( x \):
\[ y = \sqrt{2} \cos 2x + 4 \sin x \]
\[ \frac{dy}{dx} = \sqrt{2} \cdot (-2 \sin 2x) + 4 \cos x \]
\[ \frac{dy}{dx} = -2\sqrt{2} \sin 2x + 4 \cos x \]

2. Solve for \( \frac{dy}{dx} = 0 \):
\[ -2\sqrt{2} \sin 2x + 4 \cos x = 0 \]
\[ 2\sqrt{2} \sin 2x = 4 \cos x \]
\[ \sin 2x = 2 \cos x / \sqrt{2} \]
\[ \sin 2x = \sqrt{2} \cos x \]
\[ \sin 2x = \sqrt{2} \cos x \]

Using the double angle identity \(\sin 2x = 2 \sin x \cos x\):
\[ 2 \sin x \cos x = \sqrt{2} \cos x \]

If \( \cos x \neq 0 \):
\[ 2 \sin x = \sqrt{2} \]
\[ \sin x = \frac{\sqrt{2}}{2} \]
\[ x = \frac{\pi}{4} \]

Next, consider \(\cos x = 0\) within the interval \([0, \pi]\):
\[ x = \frac{\pi}{2} \]

3. Evaluate the function at the critical points and at the endpoints \( x = 0 \) and \( x = \pi \):

At \( x = 0 \):
\[ y(0) = \sqrt{2} \cos 0 + 4 \sin 0 = \sqrt{2} \]

At \( x = \pi \):
\[ y( \pi ) = \sqrt{2} \cos (2 \pi) + 4 \sin (\pi) = \sqrt{2} (1) + 0 = \sqrt{2} \]

At \( x = \frac{\pi}{4} \):
\[ y \left( \frac{\pi}{4} \right) = \sqrt{2} \cos \left( \frac{\pi}{2} \right) + 4 \sin \left( \frac{\pi}{4} \right) \]
\[ = \sqrt{2} \cdot 0 + 4 \cdot \frac{\sqrt{2}}{2} \]
\[ = 2\sqrt{2} \]

At \( x = \frac{\pi}{2} \):
\[ y \left( \frac{\pi}{2} \right) = \sqrt{2} \cos \left( \pi \right) + 4 \sin \left( \frac{\pi}{2} \right) \]
\[ = \sqrt{2} \cdot (-1) + 4 \cdot 1 \]
\[ = -\sqrt{2} + 4 \]

4. Compare the values:

\[ y(0) = \sqrt{2} \]
\[ y \left( \frac{\pi}{4} \right) = 2\sqrt{2} \]
\[ y \left( \frac{\pi}{2} \right) = 4 - \sqrt{2} \]
\[ y( \pi ) = \sqrt{2} \]

Thus, the maximum value (GTLN) is \( y \left( \frac{\pi}{4} \right) = 2\sqrt{2} \) and the minimum value (GTNN) is \( y \left( \frac{\pi}{2} \right) = 4 - \sqrt{2} \).