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Let's work through each of the three exercises one by one.

### Exercise 1:
The first problem is:
\[
\sin \alpha + \cos \alpha = \frac{5}{3}, \quad \text{calculate} \quad 2\sin \alpha \cdot \cos \alpha.
\]
This uses the identity for the sine of a double angle:
\[
2\sin \alpha \cdot \cos \alpha = \sin(2\alpha).
\]
We will also use the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\).

Let's begin by solving for \(\sin \alpha\) and \(\cos \alpha\).

### Exercise 2:
The second problem gives the equation:
\[
\frac{6\sin \alpha + 7\cos \alpha}{\cos \alpha} = 4, \quad \text{calculate} \quad \tan \alpha.
\]
Simplify the expression and solve for \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\).

### Exercise 3:
In the third problem, it is given:
\[
\sin \alpha = \frac{2\sqrt{7}}{7}, \quad \text{calculate} \quad \cos^2 \alpha - 2.
\]
We can use the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\) to first find \(\cos^2 \alpha\).

I'll start solving these step by step.Here are the results for each exercise:

### Exercise 1:
Given \(\sin \alpha + \cos \alpha = \frac{5}{3}\), the value of \(2\sin \alpha \cdot \cos \alpha\) is approximately:
\[
2\sin \alpha \cdot \cos \alpha = 1.78.
\]

### Exercise 2:
The equation \(\frac{6\sin \alpha + 7\cos \alpha}{\cos \alpha} = 4\) does not produce any real solutions for \(\tan \alpha\), suggesting no real solution exists for this setup.

### Exercise 3:
Given \(\sin \alpha = \frac{2\sqrt{7}}{7}\), the value of \(\cos^2 \alpha - 2\) is:
\[
\cos^2 \alpha - 2 = -\frac{11}{7}.
\]

If you'd like more details or have any questions, let me know!

Here are 5 related questions:
1. What other trigonometric identities could be useful for solving these problems?
2. How can we determine if an equation has no real solutions?
3. What is the geometric interpretation of these trigonometric results?
4. How would the answers change if \(\alpha\) was an obtuse angle instead of acute?
5. How can we generalize these trigonometric problems to other angles?

**Tip**: When solving trigonometric equations, always check for extraneous solutions or restrictions based on the angle's range.

Kąt α jest kątem ostrym i sina + cosa = 5/3, oblicz 2sina * cosa. Kąt α jest kątem ostrym i (6sina + 7cosa)/cosa = 4, oblicz tga. Kąt α jest kątem ostrym i sina = (2√7)/7, oblicz cosa^2 - 2.

Solve trigonometric problems involving sine, cosine, and tangent functions. This set of exercises covers acute angles, trigonometric identities, and step-by-step solutions for calculating values like 2sinα * cosα and tanα. Suitable for high school students.

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