The embeddings $S^{6} \to S^{7}$ and $S^{7}\to S^{8}$ Is the standard smooth structure of $S^{7}$ the only structure for which each of the following canonical  embedding is  an smooth embedding?:
1)$S^{6}\to S^{7}$
2)$S^{7}\to S^{8}$
 A: I will interpret (the negation of) your questions in the following way.

Let $n>5$ and consider the topological embedding of the equator $S^{n-1}$ in a topological $S^n$. 
  1) Is there an exotic smooth structure on $S^n$ such that, for the round structure on $S^{n-1}$ the embedding is smooth? 
  2) Does any exotic smooth $S^{n-1}$ embed smoothly in the standard $S^{n}$?

These kind of problems are usually solved by the $h$-cobordism theorem and its applications.
1) Yes (that is, your first question has a negative answer).
Consider a smooth structure $X$ on $S^n$ and let $S^{n-1}\hookrightarrow X$ be a smooth embedding that bounds a standard ball $B$ in a chart $U$ for $X$. There is a homeomorphism $X\to S^n$ that restricts to a diffeomorphism of $B$ onto the lower hemisphere (e.g. because $X$ is a twisted sphere). This shows that you can map $S^{n-1}$ into any exotic $S^n$ as the "equator".
2) No (that is, your second question has a positive answer).
This is a generalisation of the Schoenflies problem: if $S^{n-1}$ is a smooth submanifold of $S^n$ for $n\ge 5$, then it's diffeomorphic to the standard $S^{n-1}$.
EDIT: Notice that, as Johannes Nordström points out in a comment below, if you fix the smooth structure on the target $S^n$ as well as the topological embedding whose image is a smooth submanifold, you automatically get a smooth structure on the source $S^{n-1}$, so the answer to your second question, as stated in your post, is yes, with no need to use any "big" result (no $h$-cobordism, no Schoenflies, etc.). In the previous version of this post, my question 2) was the negation of your original question, hence the answer was quite an overkill, while now it's more general.
