I dont know how i can resolve this problem of the ε definition of a limit of succession

For the first one, you have to prove that, given $\ \varepsilon>0,\ \exists\ N\in\mathbb{N}\ $ such that $$\ \left\vert\frac{2n+3}{3n-50} - \frac{2}{3}\right\vert\ < \varepsilon\quad \forall\ n\geq N.$$

So consider the fact that

$$\frac{2n+3}{3n-50} - \frac{2}{3} = \frac{6n+9}{9n-150} - \left(\frac{6n-100}{9n-150} \right) = \frac{109}{9n-150}\ < \frac{109}{n}\quad \forall n\geq 19.$$

Therefore, for all $\ n\geq 19,\ $

$$ 0 < \left\vert\frac{2n+3}{3n-50} - \frac{2}{3}\right\vert = \frac{2n+3}{3n-50} - \frac{2}{3} = \frac{109}{9n-150} < \frac{109}{n},$$

and you should be able to use the Archimedean property of real numbers to complete the $\ \varepsilon-n\ $ proof.

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For the second one, you must show that given $\ x\in\mathbb{R},\ \exists\ N\in\mathbb{N}\ $ such that $$\ \frac{3n^2-12n+1}{n+25} > x\quad \forall n\geq N.\ $$

To do this, just use polynomial long division and then show that you can find an $\ N\ $ that does this for each $\ x.$