Task 1. From the relation $x_{n-1}=\frac{x_{n}+x_{n-1}+x_{n-2}}{3}$, we get: $2 x_{n-1}=x_{n-2}+x_{n}$ for all $n \geq 3$. This means that the given sequence is an arithmetic progression. Let the common difference of the progression be $d$. Then $\frac{x_{300}-x_{33}}{x_{333}-x_{3}}=\frac{(300-33) d}{330 d}=\frac{89}{110}$.
Answer: $\frac{89}{110}$.