$C^k(M)$ is isomorphic to $C(M)$ and hence, by Milutin's Theorem, to $C[0,1]$. The first statement is more or less obvious since the norm on $C^k(M)$ is equivalent on a $k$-codimensional subspace to the sup norm of the $k$-th derivative and $C(M)$ contains a (necessarily complemented) subspace isomorphic to $c_0$.

I don't remember the definition of strongly regular and don't have the Ghoussoub et al Memoir here to look it up.

$C^k(M)$ is isomorphic to $C(M)$ and hence, by Milutin's Theorem, to $C[0,1]$. The first statement is more or less obvious since the norm on $C^k(M)$ is equivalent on a $k$-codimensional subspace to the sup norm of the $k$-th derivative and $C(M)$ contains a (necessarily complemented) subspace isomorphic to $c_0$.

I don't remember the definition of strongly regular and don't have the Ghoussoub et al Memoir here to look it up.

EDIT 3/6/17: As Mikhail Ostrovskii mentioned in a comment, $C^k(M)$ is NOT isomorphic to $C(M)$ when the dimension of $M$ is two or more, at least for some $M$.