I want to answer the following question: $$\vec\nabla \times (\vec a \times \vec F) = \vec a(\vec \nabla \cdot \vec F)-(\vec a \cdot \vec\nabla)\vec F.$$ 'By applying Stokes' theorem to the above identity, show that $$\int_C d\vec l \times \vec F = \int_S (d\vec S\times \vec \nabla)\times \vec F$$ for any closed curve $C$ that bounds a surface $S$.'
I have no idea what is meant by $d\vec S \times \vec \nabla$ or how to approach this question; I understand Stokes' theorem and what is meant by $(\vec a \cdot \vec \nabla)\vec F$, but I don't think you are allowed to use standard properties of the cross product (anticommutative etc.) to rearrange the RHS.
$\vec a$ and $\vec F$ are generic vector fields.
Here, $d\vec{l}$ is the (vectorial)-length element along the curve and $d \vec S$ the area element where the vector points orthogonal to the element.
If you apply Stokes to the first equation, you obtain $$\int_{C}d\vec{l}\cdot(\vec{a}\times \vec{F}) = \int_S d\vec S \cdot\left[\vec a(\vec \nabla \cdot \vec F)-(\vec a \cdot \vec\nabla)\vec F\right].$$
Next, we employ the triple product identity $$ d\vec{l}\cdot(\vec{a}\times \vec{F}) = -\vec{a}\cdot(d\vec{l}\times \vec{F}),$$ the relation $$d\vec S \cdot\left[\vec a(\vec \nabla \cdot \vec F)-(\vec a \cdot \vec\nabla)\vec F\right]= \vec a \cdot\left[ d\vec S (\vec \nabla \cdot \vec F) - \sum_{j=1}^3 dS_j \vec{\nabla} F_j\right] ,$$ and Lagrange's formula (were we have to make sure that the derivative acts only on $\vec{F}$) $$(d\vec{S}\times \vec{\nabla})\times \vec{F} = -d\vec S (\vec\nabla\cdot \vec F) + \sum_{j=1}^3 dS_j \vec\nabla F_j.$$
Putting everything together, we can conclude that $$\vec a \cdot \int_C d\vec l \times \vec F = \vec a \cdot \int_S (d\vec S \times \vec\nabla)\times \vec F.$$ As $\vec a $ is arbitrary, we have shown the result as requested.