The following functions are written on the board, $$F(x) = x^2 + \frac{12}{x^2}, G(x) = \sin(\pi x^2), H(x) = 1.$$ If functions $f,g$ are currently on the board, we may write on the board the functions $$f(x) + g(x), f(x) - g(x), f(x)g(x), cf(x)$$ (the last for any real number $c$). Can a function $h(x)$ appear on the board such that $$|h(x) - x| < \frac{1}{3}$$ for all $x \in [1,10]$ ?
1. **Initial Setup and Functions on the Board:**
   We start with the functions \( F(x) = x^2 + \frac{12}{x^2} \), \( G(x) = \sin(\pi x^2) \), and \( H(x) = 1 \). We are allowed to write new functions on the board using the operations \( f(x) + g(x) \), \( f(x) - g(x) \), \( f(x)g(x) \), and \( cf(x) \) for any real number \( c \).

2. **Objective:**
   We need to determine if there exists a function \( h(x) \) that can be written on the board such that \( |h(x) - x| < \frac{1}{3} \) for all \( x \in [1, 10] \).

3. **Strategy:**
   We will focus on the functions \( F(x) \) and \( H(x) \) and ignore \( G(x) \). We aim to construct a function \( h(x) \) such that \( |h(x) - x| < \frac{1}{2} \) for all \( x \in [\sqrt[4]{12}, 10] \).

4. **Injectivity of \( F(x) \):**
   The function \( F(x) \) is injective over the interval \( [\sqrt[4]{12}, 10] \). This can be shown by solving the equation \( x^2 + \frac{12}{x^2} = y^2 + \frac{12}{y^2} \):
   \[
   x^2 + \frac{12}{x^2} = y^2 + \frac{12}{y^2} \implies (x^2 - y^2)(12 - x^2 y^2) = 0.
   \]
   Since \( x \neq y \) and \( x, y \in [\sqrt[4]{12}, 10] \), the only solution is \( x = y \).

5. **Constructing \( h(x) \):**
   Since \( F \) is injective, we can write functions of the form \( P \circ F(x) \), where \( P \) is a real polynomial. By the Stone-Weierstrass Theorem, we can approximate any continuous function on a compact interval with polynomials. Therefore, we can construct a polynomial \( P \) such that \( |P(F(x)) - x| < \frac{1}{2} \) for all \( x \in [\sqrt[4]{12}, 10] \).

6. **Extending \( h(x) \) to \( [1, \sqrt[4]{12}) \):**
   Since \( F(x) = F\left(\frac{\sqrt{12}}{x}\right) \), we have \( h(x) = h\left(\frac{\sqrt{12}}{x}\right) \). For \( x \in [1, \sqrt[4]{12}) \), we note that \( \frac{\sqrt{12}}{x} \in [\sqrt[4]{12}, 10] \). Thus,
   \[
   |h(x) - x| \leq \left|h(x) - \frac{\sqrt{12}}{x}\right| + \left|\frac{\sqrt{12}}{x} - x\right| = \left|h\left(\frac{\sqrt{12}}{x}\right) - \frac{\sqrt{12}}{x}\right| + \left|\frac{\sqrt{12}}{x} - x\right| < \frac{1}{2} + \left|\frac{\sqrt{12}}{x} - x\right|.
   \]
   Since \( \left|\frac{\sqrt{12}}{x} - x\right| \leq \sqrt{12} - 1 \) for \( x \in [1, \sqrt[4]{12}) \), we have
   \[
   |h(x) - x| < \frac{1}{2} + (\sqrt{12} - 1) < 3.
   \]

7. **Conclusion:**
   The function \( h(x) \) constructed satisfies \( |h(x) - x| < \frac{1}{3} \) for all \( x \in [1, 10] \).

The final answer is \( \boxed{ h(x) } \) such that \( |h(x) - x| < \frac{1}{3} \).