I have equation given like this: $y(t)$ = $\beta_0$ + 0,5($y(t-1)$+$y(t-2)$)+$u(t)$, $u(t)$ ~ iiN(0,sigma^2)
In AR(1) we look at the $\beta_1$ before $y(t-1)$ and if its $|\beta_1|<1$ its covariance stationary (or maybe I don't understand it right) but, How to check if this process is covariance stationary? What are the easiest ways to check it in this kind of models? Is the way to check covariance stationarity of AR(2) the same for AR(m)
First of all, the $AR(1)$ equation $$ y_t = \beta_1 y_{t-1} + u_t \text{ with } u_t \sim WN(0,\sigma^2) $$ has a covariance (weakly) stationary solution even when $| \beta_1 | > 1$ (see for example the first answer here). The reason why this solution is usually excluded is that it is non-causal (future dependent). So, to be precise, the only requirement for an $AR(1)$ equation to have a covariance stationary solution is that $|\beta_1| \neq 1$ (the stronger requirement $|\beta_1| < 1$ is needed to also ensure causality).
For a general $AR(p)$ process $$ y_t = \sum_{i = 1}^p \beta_i y_{t-i} + u_t \text{ with } u_t \sim WN(0, \sigma^2)$$ one can show that it is covariance stationary, if its autoregressive polynomial $$ \Phi(z) = 1 - \sum_{i = 1}^p \beta_i z^i $$ has no zero $z \in \mathbb{C}$ on the unite circle (i.e. $\Phi(z) \neq 0$ for every $|z| = 1$). Notice how this is equivalent to $|\beta_1| \neq 1$ for the case $p = 1$: for the $AR(1)$ process defined above we have $\Phi(z) = 1-\beta_1 z$, whose unique root is $z = 1/\beta_1$.
For the $AR(p)$ process $y(t)$ to be causal, we also need to require that the autoregressive polynomial $\Phi(z)$ does not have any root inside the unit circle (i.e. $\Phi(z) \neq 0$ for every $|z| < 1$). In the $AR(1)$ case, this is equivalent to $|1 / \beta_1 | > 1$, that is $|\beta_1| < 1$.
Let us now look at your specific example: $$y_t = \beta_0 + 0.5 y_{t-1} + 0.5 y_{t-2} + u_t.$$ We first deal with the $\beta_0$ term. Taking expectations on both sides, we have $$ \mathbb{E}y_t = \beta_0 + 0.5\mathbb{E}y_{t-1} + 0.5 \mathbb{E} y_{t-2} + 0.$$ Knowing that the process is covariance stationary only if $\mathbb{E}y_t$ does not depend on time $t$, and letting $\mu = \mathbb{E}y_t$, we obtain $\mu = \beta_0 + \mu$, which is only possible if $\beta_0 = 0$. We can thus ignore the $\beta_0$ and rewrite the process in the (standard) $AR(2)$ form: $$ y_t = 0.5 y_{t-1} + 0.5 y_{t-2} + u(t).$$ The autoregressive polynomial is $\Phi(z) = 1 - 0.5 z - 0.5 z^2$ which has two roots: $1$ and $-2$. Since one of the roots lies on the unit circle, we deduce that the process is not covariance stationary.