I'm studing cohomology with $\mathbb{Z}_2$ coefficients in order to understand Stiefel-Whitney classes, and i have a couple of questions.Let $X$ be a n-manifold.

1) I know that $H^k_{sing}(X,\mathbb{Z})\subset H^k_{sing}(X,\mathbb{R})\simeq H^k_{DR}(X)$, and that the wedge product $\wedge$ is the cup product $\cup$ for $H^k_{DR}(X)$. Thank to the mentioned inclusion, i can use the wedge product also on $H^k_{sing}(X,\mathbb{Z})$. Now i am wondering what happens for singular cohomology with $\mathbb{Z_2}$ coefficients. A priori no wedge product is defined, but using the Universal coefficient theorem, if $Ext (H_{k-1}(X,\mathbb{Z}),\mathbb{Z}_2)=0$ then every element in $H^k(X,\mathbb{Z_2})$ is the reduction mod 2 of an element in $H^k(X,\mathbb{Z})$, but on $H^k(X,\mathbb{Z})$ the wedge product is defined, while on $H^k(X,\mathbb{Z_2})$ is not. how does it work? when i reduce mod 2 the wedge product becomes the cup product?

2) Given $\omega \in H^k(X,\mathbb{R})$ and $[S]$ a k-dimensional cycle i know we have the pairing $\langle [S],\omega\rangle:= \int_s\omega$. How is this defined for $H^k(X,\mathbb{Z}_2)$? Apparently, when $\omega \in H^k(X,\mathbb{Z}_2)$ is the reduction mod 2 of $\widetilde{w}\in H^k(X,\mathbb{Z})$, it is ok to set $\langle [S],\omega\rangle=\int_S \widetilde{\omega}$ mod 2. For example in the proof of the wu's formula http://books.google.it/books?id=nqMNrHE1U28C&pg=PA163&lpg=PA163&dq=%22wu's+formula%22&source=bl&ots=ksKOp7Hw6C&sig=8GQt7P48siLl8YGnqSI4HRQjckk&hl=it&sa=X&ei=DrtAUeD2KoSq4AS4vYCIBg&ved=0CC8Q6AEwAA#v=onepage&q=%22wu's%20formula%22&f=false (here the notation is $w_2(T_M)\cdot S$) Scorpan says that $w_2(T_S) $is the reduction mod 2 of the euler class, but the integral must be implicit because $\int_S e(T_S)=\mathcal{X}(S)$, but is the integral defined for elements in $H^k(X,\mathbb{Z_2})$? How does it work?

Thank you