TITLE: Orientation and generalized cohomology theories
QUESTION [12 upvotes]: Let h* be a multiplicative generalized cohomology theory and $E \rightarrow X$ a real vector bundle.

Is it true that, if $E$ is orientable with respect to h*, then it is also orientable with respect to the singular cohomology with coefficients in $h^{0}(pt)$, for $(pt)$ a space with one point?
Is it true that, if $E$ is orientable with respect to the singular cohomology with coefficeints in $\mathbb{Z}_{p}$, for $p > 2$, then it is also orientable with respect to the singular cohomology with coefficients in $\mathbb{Z}$?

Added later: I thought to a possible simple answer using weak orientability, actually proving what stated in the last comment. I wrote this as a comment to answer 3.

REPLY [3 votes]: At the risk of being obvious, let me make some general comments:
For a vector bundle $E\to X$ of rank $n$ let $E_x$ be the fiber over $x\in X$. For a multiplicative cohomology theory $h$, the cohomology $h^*(E_x,E_x-0)$ is a free module of rank $1$ over the graded ring $h^*$, with generator in $h^n$. Such a generator is called an orientation at the point $x$. A Thom class (also called an orientation) is an element of $h^n(E,E-0)$ that restricts to such a generator for every $x$.
There is also the other sense of orientation: a continuous choice of pointwise orientation for all points. In the special case of ordinary cohomology, in other words when the graded ring $h^*(point)$ is all in degree zero, an orientation in this weak sense implies an orientation in the Thom-class sense; but not in general.