Is there an infinity of primes $p$ with the property $p\equiv2\pmod3$, and how is it proven?
Modulo $3$, there is only one number who's congruent to $0$, it is $3$ itself. But that is pretty obvious. But we have $2\equiv2\pmod3$, $5\equiv2$, $11\equiv2$, $17\equiv2$,... But is there an infinity?
Yup. Suppose I have a list of primes $p_1, . . . , p_n$ which equal $2$ mod $3$. Then consider $n=p_1^2p_2^2\cdot\cdot\cdot p_n^2$. Since $2^2=1$ mod $3$, we have $n=1$ mod $3$. So now think about $n+1$.
$n+1$ is $2$ mod $3$, so its prime factors don't include $3$ and can't all be $1$ mod $3$; that is, $n+1$ must be divisible by some prime $q$ which is $2$ mod $3$. But $p_i\not\mid n+1$ (why?), so $q\not=p_1, p_2, . . . , p_n$.
See also Dirichlet's theorem.