Prove that any differential form $w\in S^1$ is cohomologous to a form which is invariant with respect to rotation
I don't understand, what means "form, which invariant with respect to rotation". If we have an action of a circle on itself $M:S^1 \to S^1$ by rotation, $M(x)=x+M$, so how to define the differential form $M(w)$? Then I should show that there $f$ such that $w-M(w)=df$, but how to find such $f$?
If $f: M \to N$ is a smooth function, it induces a pull-back map $f^*: \Omega^k(N) \to \Omega^k(M)$. In your case, if $r: \mathbb{S}^1 \to \mathbb{S}^1$ is a rotation map, then a rotation invariant form $\omega \in \Omega^k(\mathbb{S}^1)$ is one such that $r^*(\omega) = \omega$. Therefore, you are asked to find, for any form $w \in \Omega^1(\mathbb{S}^1)$ a form $\omega$ such that $r^*(\omega) = \omega$, and there exists a function $f \in \Omega^0(\mathbb{S}^1) = C^\infty(\mathbb{S}^1)$ such that $w - \omega = d f$.