Claim 1: If $R$ is a principal ideal domain, then there is a short exact sequence $$0 \to Ext^1_R(H_{n-1}(X;R),R) \to H^n(X;R) \xrightarrow[]{\tilde{h}} Hom_R(H_n(X;R),R) \to 0\hspace{10pt}(\ast)$$
This answers b) and c).
Claim 2: If $R$ or $H_\ast(X)$ is torsion-free, then $\tilde{h} = h$ (for $G=R$).
Proof: Denote the simplicial complex of $X$ by $S(X)$. Then $C := S(X) \otimes_{\mathbb Z} R$ is a complex of free $R$-modules. Hence, by the universal coefficient theorem (cf. Hatcher, after Corollary 3.4) we have the short exact sequence $$0 \to Ext^1_R(H_{n-1}(C),R) \to H^n Hom_R(C,R) \xrightarrow[]{\tilde{h}} Hom_R(H_n(C),R)\to 0.$$ But $Hom_R(C,R) = Hom_R(S(X) \otimes_{\mathbb Z} R,R) \cong Hom(S(X),R)$ (cf. III (3.3) in Brown: Cohomology of Groups) and $H_n(C) = H_n(X;R)$. Hence the short exact sequence above transforms into $(\ast)$.
Moreover, by universal coefficients, $$H_n(X;R) = H_n(X) \otimes_{\mathbb Z}R \oplus Tor_1^{\mathbb Z}(H_{n-1}(X),R) = H_n(X) \otimes_{\mathbb Z}R$$ if $R$ or $H_{n-1}(X)$ is torsion-free. Hence $$Hom_R(H_n(X;R),R) = Hom_R(H_n(X) \otimes_{\mathbb Z} R,R) \cong Hom(H_n(X),R).$$
This shows claim 2.
Remark: a) Example 3.8 from Hatcher uses $R=\mathbb{Z}/2$.
b) Example 3.9 uses $R=\mathbb{Z}/m$. Hence $(\ast)$ doesn't apply (for general $m$). In particular it can't be used to show that $\tilde{h}$ is surjective. This may be the reason for Hatcher's alternative argumentation in this example.