I hope I write down (a) correctly. For (b), I followed Amitesh Datta's suggestion, but I hope I well-justified my argument - did I?
On $\mathbb{R}^2$, let $\omega = (\sin^4 \pi x + \sin^2 \pi(x + y))dx - \cos^2 \pi(x + y)dy$.
(a) Show that $\omega$ is closed.
By definition, a $p$-form $\omega$ on $X$ is closed if $d\omega=0$. So \begin{align*} d\omega & = \frac{\partial}{\partial y}(\sin^4 \pi x + \sin^2 \pi(x + y)) \wedge dx - \frac{\partial}{\partial x} \cos^2 \pi(x + y) \wedge dy\\ & = 2 \pi \sin \pi(x + y)) \cos \pi (x+y) dy \wedge dx + 2 \pi \cos \pi(x + y) \sin \pi(x + y) dx \wedge dy\\ &= 0 \end{align*}
(b) Let $\eta$ be the unique $1$-form on the torus $T^2 = \mathbb{R}^2 / \mathbb{Z}^2$ such that $p^* \eta = \omega,$ where $p: \mathbb{R}^2 \to T^2$ is projection. The parametrized curve $\gamma: \mathbb{R} \to \mathbb{R}^2$ given by $\gamma(\theta) = (2\theta, -3\theta)$ is a line whose image $C \subset T^2$ is an oriented circle. Prove that $\int_C \eta \neq 0$. What is its sign?
By the theorem
Change of Variable in $\mathbb{R}^k$. Assume that $f: V \to U$ is an orientation-preserving diffeomorphism of open sets in $\mathbb{R}^k$ or $\mathbb{H}^k$, and let $\omega$ be an integrable $k$-form on $U$. Then $$\int_U \omega = \int_V f^*\omega.$$ If $f$ reverses orientation, then $$\int_U \omega = - \int_V f^* \omega.$$
So for this problem, we have $p: \mathbb{R}^2 \to \mathbb{T}^2$, and I believe projection map is orientation preserving, and I know $p$ is diffeomorphism of open sets in $\mathbb{R}^k$. And I granted $\omega$ is integrable for it is composed of Trigonometric functions. Then I reached my conclusion that:
$$\int_C \eta = \int_\gamma p^* \eta = \int_\gamma \omega = \int_\gamma (\sin^4 \pi x + \sin^2 \pi(x + y))dx - \cos^2 \pi(x + y)dy.$$ Carry out line integrals with $\gamma(\theta) = (2\theta, -3\theta):$ $$\int_\theta (\sin^4 \pi 2\theta + \sin^2 \pi \theta) 2d\theta + \cos^2 \pi \theta 3d \theta.$$
Hence, as long as $\theta \not\equiv 0$, $\int_C \eta \neq 0$, and it is positive.
Hint: $\int_{C} \eta=\int_{\gamma} \omega$