Let $\mathfrak{g}$ be a complex semi-simple Lie algebra. Let $\mathfrak{h}$ be a cartan subalgebra. Let $ \Delta $ be the resulting root system. Denote by $ V $ the real span of the roots. Let $ \alpha_1,...,\alpha_r$ be simple roots and $ \check{\alpha_i} $ the corresponding coroots. Let $ W $ be the Weyl group. Recall that it generated by the reflections $ s_i:=s_{\alpha_i}$. $ W $ can also be seen on the Lie group level as $N_G(T)/T $. Denote by $ S \mathfrak{h} $ the symmetric algebra on $ \mathfrak{h} $. It can be a seen as a polynomial ring on $ \mathfrak{h}^*$.
Let $ P \in S \mathfrak{h} $, $w \in W$ and $ H \in \mathfrak{h}^* $. My questions are the following.
1) Is it correct that $ w $ act on $ P $ as $ (w.P)(H)=P(w^{-1}H) $ ?
2) How does $ W $ act on $ H \in \mathfrak{h}^* $ if $ H $ isn't in $ V $ ?
3) Is it true that $ s_i \check{\alpha_j} = \check{s_i \alpha_j}$ ? Edit: after some computations, I think that this is only true iff the Cartan matrix is symmetric. In particular it is true for types A,D,E.
For question one, that's certainly a way for $W$ to act on $P$. Whether or not it is correct depends on the context.
2): The roots are a subset of $\mathfrak{h}^*$, and the Weyl group acts on those roots. So you extend that action by linearity. Whether you extend to real scalars or complex ones (or whatever) doesn't really make a difference. Do you maybe mean to ask how it acts on elements of $S\mathfrak{h}$ that aren't in $\mathfrak{h}$? If so, the answer is that if $h_1\cdots h_k$ is a monomial with each $h_i \in \mathfrak{h}$ and $w\in W$ then $w\cdot (h_1\cdots h_k) = (w\cdot h_1)(w\cdot h_2)\cdots(w\cdot h_k)$.
3) Yes: use that $\check{\alpha} = 2\alpha/(\alpha,\alpha)$ and the fact that $W$ preserves the inner product, i.e. $(\alpha,\alpha) = (s\alpha, s\alpha)$ for any $s\in W$.