Show that $(\iota^*\omega)_p(v_1,v_2) =1$ 
Let $\omega$ be the $2$-form on $\Bbb R^3\setminus\{0\}$ defined by
$$\omega = (x^2+y^2+z^2)^{-3/2}(xdy\wedge dz-ydx\wedge dz+zdx\wedge dy).$$
$(1)$ Let $\iota:\Bbb S^2\to\Bbb R^3$ be the inclusion map and consider $T_p\Bbb R\cong\Bbb R^3$ with Euclidean inner product. If $v_1,v_2$ are positively oriented orthonormal basis of $T_p\Bbb S^2\subset T_p\Bbb R^3$ then $(\iota^*\omega)(v_1,v_2) =1$.
$(2)$ $\omega$ is closed but not exact on $\Bbb R^3\setminus\{0\}$.

My attempt: (1) Let $p =(x,y,z)$ then
$$(\iota^*\omega)_p = xdy_p\wedge dz_p-ydx_p\wedge dz_p+zdx_p\wedge dy_p.$$
Write
\begin{align*}
v_1 & = a{\partial\over\partial x}\bigg|_p+b{\partial\over\partial y}\bigg|_p+c{\partial\over\partial z}\bigg|_p\\
v_2 & = d{\partial\over\partial x}\bigg|_p+e{\partial\over\partial y}\bigg|_p+f{\partial\over\partial z}\bigg|_p\\
\end{align*}
for some $a,...,f\in\Bbb R$. Since $v_1,v_2$ are positively oriented orthonormal basis, $p\cdot (v_1\times v_2) =1$. Hence by computation,
$$bfx-afy+aez+(-ecx+dcy-bdz) =1.$$
Note that
\begin{align*}
(\iota^*\omega)_p(v_1,v_2) & = bfx-afy+aez\\
-(\iota^*\omega)_p(v_2,v_1) & = -ecx+dcy-bdz\\
\end{align*}
As an alternating form, $-(\iota^*\omega)_p(v_2,v_1) = (\iota^*\omega)_p(v_1,v_2)$. Hence,
$$(\iota^*\omega)_p(v_1,v_2) = {1\over 2}.$$
(2) Routine computation shows $d\omega =0$ so $\omega$ is closed. Now suppose $\omega = d\eta$ for some $1$-form $\eta$ on $\Bbb R^3\setminus\{0\}$. Then $\iota^*\omega$ is a $2$-form on $\Bbb S^2$ so $\iota^*\omega =\iota^*d\eta = d\iota^*\eta$ and $\iota^*\eta$ is a $1$-form on $\Bbb S^2$. By Stoke's theorem
$$\int_{\Bbb S^2}\iota^*\omega = \int_{\Bbb S^2}d\iota^*\eta = \int_{\partial\Bbb S^2}\iota^*\eta =0,$$
as $\partial\Bbb S^2 = \emptyset$. But by (1), we know $\iota^*\omega$ is an orientation form, i.e., smooth nonvanishing $2$-form on $\Bbb S^2$. In particular, $\int_{\Bbb S^2}\iota^*\omega >0$ which is a contradiction. So, not exact.
Something is wrong in (1). Please help.
 A: Your error was stating that $\alpha \wedge \beta (v,w) = \alpha(v)\beta(w)$ for 1-forms $\alpha,\beta$.
It is $\alpha(v)\beta(w) -\beta(v)\alpha(w)$, or in a more condensed form, $\begin{vmatrix} \alpha(v) & \alpha(w) \\ \beta(v) & \beta(w)\end{vmatrix}$.
If $v_1,v_2$ is a positively oriented orthonormal basis of $T_pS^2$, then $\{p,v_1,v_2\}$ is a positively oriented orthonormal basis of $\Bbb R^3$ (recall that $T_pS^1=\{p\}^{\perp}$).
It follows that if $p=(x,y,z)$, then
\begin{align}
1 &= \det(p,v_1,v_2)\\
&=\begin{vmatrix}x & dx(v_1) & dx (v_2) \\ y & dy(v_1) & dy(v_2) \\ z & dz(v_1) & dz(v_2)\end{vmatrix} \\
&= x \begin{vmatrix}dy(v_1) & dy(v_2) \\ dz(v_1) & dz(v_2)  \end{vmatrix} - y  \begin{vmatrix}dx(v_1) & dx(v_2) \\ dz(v_1) & dz(v_2)  \end{vmatrix} + z \begin{vmatrix}dx(v_1) & dx(v_2) \\ dy(v_1) & dy(v_2)  \end{vmatrix} \\
&= (x dy\wedge dz - y dx\wedge dz + z dx\wedge dy)(v_1,v_2)\\
&= (\iota^*\omega)_p(v_1,v_2),
\end{align}
which is the desired result.
More generally, the $n-1$-form $\omega_p(v_1,\ldots,v_{n-1}) = \det (p,v_1,\ldots,v_{n-1})$ is a volume form on $S^{n-1}\subset \Bbb R^n$.
