Let's analyze the mathematical statements in the uploaded image one by one.

1. На интервале $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$ найдется точка $c$, в которой производная функции $y = \cos x$ равна нулю.

• The derivative of $y = \cos x$ is $y' = -\sin x$.
• For $y' = 0$, we solve $-\sin x = 0$, which means $\sin x = 0$.
• Within the interval $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$, $\sin x$ is never zero because the sine function is negative in this range.

Conclusion: This statement is false.

2. Точка $x_0 = 0$ — точка строгого локального минимума функции $y = |x|$.

• The function $y = |x|$ is defined as: x, & \text{if } x \geq 0, \\ -x, & \text{if } x < 0. \end{cases}$$
• At $x_0 = 0$, the value of $y$ is 0, and it is the lowest point of the function. Since $|x|$ is symmetric and increasing in both directions, $x_0 = 0$ is a strict local minimum.

Conclusion: This statement is true.

3. Поскольку функция $y = |x|$ в точке $x_0 = 0$ принимает экстремальное значение и имеет конечную производную $y'(x_0)$, то $y'(x_0) = 0$.

• At $x = 0$, $y = |x|$ has a cusp (corner point), meaning the derivative is undefined there. The left-hand derivative is $-1$, and the right-hand derivative is $1$. Therefore, $y'(x_0)$ does not exist.

Conclusion: This statement is false.

4. На интервале $(0, 2)$ найдется точка $c$, в которой производная функции $y = x^3$ равна $y'(c) = -4$.

• The derivative of $y = x^3$ is $y' = 3x^2$. To check if $y'(c) = -4$, solve $3c^2 = -4$. However, $c^2$ cannot be negative, so there is no solution.

Conclusion: This statement is false.

5. На интервале $(0, 2)$ найдется точка $c$, в которой для производных функций $f(x) = x^4$ и $g(x) = x$ выполняется соотношение: $f'(c) = 8g'(c)$.

• The derivative of $f(x) = x^4$ is $f'(x) = 4x^3$, and the derivative of $g(x) = x$ is $g'(x) = 1$.
• For the condition $f'(c) = 8g'(c)$, we solve $4c^3 = 8$, which gives $c^3 = 2$, so $c = \sqrt[3]{2}$.
• Since $\sqrt[3]{2} \in (0, 2)$, such a point exists.

Conclusion: This statement is true.

Final Answer:

The correct statements are:

• Statement 2: $x_0 = 0$ is a strict local minimum of $y = |x|$.
• Statement 5: There exists a $c$ in $(0, 2)$ where $f'(c) = 8g'(c)$ is satisfied.

Would you like a step-by-step solution for any of these?

Here are some related questions to deepen your understanding:

1. Why does the derivative not exist at sharp points like $y = |x|$ at $x = 0$?
2. How do critical points relate to the derivative of trigonometric functions?
3. Can cubic functions have negative derivative values, and why not in this case?
4. How do higher-order derivatives influence the geometry of functions?
5. Could $c$ exist in other intervals for statement 5, and why?

Tip: Always evaluate intervals carefully when solving derivative equations, as the range of validity might exclude solutions!