Does the first statement imply the second? (a) The number $\sqrt{2}$ is either rational or irrational. If $\sqrt{2}$ rational then $\sqrt{(2)}^2$ is rational too. Either $\sqrt{2}$, or $\sqrt{(2)}^2$ is not rational.
(b) Consequently $\sqrt{2}$ is irrational.
I get this statement and I should show that this is true or false. I think I should formalize it somehow, but I still don't know how it works. But I think that the first statement doesn't imply the second one.
 A: If by first statement you mean (a), then you're correct: no, (a) does not imply (b).
Actually that last part of (a) saying "either $\sqrt{2}$, or $\sqrt{(2)}^2$ is not rational"
makes no sense (doesn't follow from anything said so far).
The first two parts of (a) are true (but they are true for any real number, not just for $\sqrt{2}$). 
A: To me, there are two ways we can take the either…or.
If we take it as exclusive (i.e. meaning exactly one of these two numbers is not rational), then (b) follows from (a). Indeed, suppose (b) didn't hold. Then $\sqrt2$ would be rational. But then $\sqrt{2}^2$ would also be, so $\sqrt2,\sqrt2^2$ would both be rational, contradicting the last part of (a).
If we take it as non-exclusive (i.e. meaning at least one of these two numbers is not rational), then (b) still follow, by the same argument.
The third part of (a) is pretty strange, because how should one know that at least one (or exactly one) of two numbers is not rational? Sounds like some logic puzzle.
My formal interpretation of the problem
Say $R(x)$ indicates "$x$ is rational". Then (a) is $(R(\sqrt2)\veebar\lnot R(\sqrt2))\land(R(\sqrt2)\vdash R(2))\land(\lnot R(\sqrt2)\vee\lnot R(2))$, and (b) is $\lnot R(\sqrt2)$, and you must prove $(a)\vdash(b)$.
To do this, try adding $R(\sqrt2)$ to (a). Now you are starting from $(R(\sqrt2)\veebar\lnot R(\sqrt2))\land(R(\sqrt2)\vdash R(2))\land(\lnot R(\sqrt2)\vee\lnot R(2))\land R(\sqrt2)$. Clearly, $(\lnot R(\sqrt2)\vee\lnot R(2))\land R(\sqrt2)\vdash\lnot R(2)$, but then you have $R(\sqrt2)\land(R(\sqrt2)\vdash R(2))\land(\not R(2))$, which implies $R(2)\land(\not R(2))$, a contradiction.
The interpretation ambiguity is as to whether the last part of (a) should have a $\vee$ or a $\veebar$, but even if it were $\lnot R(\sqrt2)\veebar\lnot R(2)$, the proof above would work just as well.
Try giving this to your teacher and seeing what he says. And excuse any logic symbol errors, since I never formalized logic this way.
