Remember that a ring is called hereditary, if submodules of projective modules are projective. If this holds only for finitely generated submodules, the ring is called semi-hereditary. 

**Claim 1:** If the ring $R$ is hereditary, 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. 

**Remarks:** a) Note that the universal coefficient theorem from Hatcher requires $R$ to be a PID. But it actually holds for hereditary rings (I think this can be found in Spanier, otherwise in Cartan-Eilenberg). 

b) Example 3.8 from Hatcher uses $R=\mathbb{Z}/2$ what's hereditary as a field. Example 3.9 uses $R=\mathbb{Z}/m$. Since a finite product of hereditary rings is hereditary, $R$ is hereditary if $m$ is square free. Moreover we have $Hom_R(-,R) = Hom(-,R)$. I don't know if $R$ is hereditary or semi-hereditary for general $m$.