Consider the set of all pairs $(\boldsymbol{n},\boldsymbol{v})$ of vectors in $\mathbb{R}^3$ such that $\boldsymbol{n}$ is a vector on the unit sphere centered at the origin and $\boldsymbol{v}$ is a unit vector tangent to the sphere at the point $\boldsymbol{n}.$
i. Introduce a structure of smooth manifold on this set.
ii. Prove that this manifold is diffeomorphic to the group $SO(3).$
To my understanding, this manifold is $S^2 \times S^1,$ which gives a parametrization of $SO(3),$ but it is far from being a diffeomorphism, i.e. the exercise is false: do you agree?
The exercise is fine. The manifold described in the exercise is called the unit tangent bundle of $S^2$ and it is not diffeomorphic to $S^2 \times S^1$ [one way to see this is to observe that you could produce a nowhere vanishing vector field on $S^2$ if they were diffeomorphic. This is impossible by the hairy ball theorem.]
Here's a slightly more detailed outline:
By definition the set $M$ is given as a subset of $\mathbb{R}^3 \times \mathbb{R}^3 \ni (\boldsymbol{n}, \boldsymbol{v})$ subject to the equations $$ \begin{align*} 1 & = \boldsymbol{n} \cdot \boldsymbol{n} && \boldsymbol{n} \text{ is a unit vector}\\ 1 & = \boldsymbol{v} \cdot \boldsymbol{v} && \boldsymbol{v} \text{ is a unit vector}\\ 0 & = \boldsymbol{n} \cdot \boldsymbol{v} && \boldsymbol{n} \text{ is perpendicular to }\boldsymbol{v}. \end{align*} $$ Can you use this information to turn $M$ into a manifold? (implicit functions, regular values, etc)
The map $M \to SO(3)$ given by $(\boldsymbol{n},\boldsymbol{v}) \mapsto [\boldsymbol{n},\boldsymbol{v},\boldsymbol{n} \times \boldsymbol{v}]$ is well-defined smooth and bijective. You can exhibit an explicit smooth inverse, so $M$ and $SO(3)$ are diffeomorphic.