= Solution
\b[No: the round sphere supplies a counterexample.] On the unit round $S^2\subset\mathbb R^3$, restrict the smooth one-form
$$
\theta=x_3^2\,dx_1.
$$
The <antipodal map> $A(x)=-x$ satisfies $A^*\theta=-\theta$, since $x_3^2$ is unchanged and $dx_1$ changes sign. Every unit-speed closed <geodesic> is a great circle, with $\gamma(t+\pi)=A(\gamma(t))$. Therefore its two half-circle integrals cancel:
$$
\int_0^{2\pi}\theta_{\gamma(t)}(\dot\gamma(t))\,dt
=\int_0^\pi\bigl(\theta+A^*\theta\bigr)_{\gamma(t)}(\dot\gamma(t))\,dt=0.
$$
The same holds for all repeated traversals and either orientation.
However, its <exterior derivative> is
$$
d\theta=2x_3\,dx_3\wedge dx_1,
$$
which is not zero on the sphere. For example, at $p=(0,1/\sqrt2,1/\sqrt2)$ the tangent vectors $u=(1,0,0)$ and $w=(0,1/\sqrt2,-1/\sqrt2)$ give $d\theta_p(u,w)=1$. Hence $\theta$ is not a <closed differential form>, and cannot be an <exact differential form>, because $d(dh)=0$. This is the <counterexample to geodesic one-form injectivity on the round sphere>; periodic geodesic integrals alone do not imply exactness without an additional geometric hypothesis.
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