Here are my solutions from 3 years ago when I was studying under Ken Brown as an undergraduate:
V.3.5:
The symmetric group $S_3$ on three letters has a semi-direct product representation $S_3=\mathbb{Z}_3\rtimes \mathbb{Z}_2$ where $\mathbb{Z}_2$ acts on $\mathbb{Z}_3$ by conjugation. Thus $H^*(S_3)=H^*(S_3)_{(2)}\oplus H^*(S_3)_{(3)}\cong H^*(S_3)_{(2)}\oplus H^*(\mathbb{Z}_3)^{\mathbb{Z}_2}$ by Theorem III.10.3, with $H^*(S_3)_{(2)}$ isomorphic to the set of $S_3$-invariant elements of $H^*(\mathbb{Z}_2)$. Exercise III.10.1 showed that $H^*(S_3)_{(2)}\cong \mathbb{Z}_2$, so it suffices to compute $H^*(\mathbb{Z}_3)^{\mathbb{Z}_2}$ where we know that $\mathbb{Z}_2$ acts by conjugation on $\mathbb{Z}_3$ ($1\mapsto 1$, $x\mapsto x^2$, $x^2\mapsto x$). But this action can be considered as the endomorphism $\alpha(2)$ from Exercise V.3.4 since $(1)^2=1$ and $(x)^2=x^2$ and $(x^2)^2=x^4=x$, and that exercise implies that the induced map on the $(2i)^{th}$-cohomology is multiplication by $2^i$ [we know that the cohomology is trivial in odd dimensions]. Now $2^1\equiv 2\,\text{mod}3$ and $2^2\equiv 1\,\text{mod}3$, so by multiplying both of those statements by $2$ repeatedly we see that $2^i\equiv 1\,\text{mod}3$ for $i$ even and $2^i\equiv 2\,\text{mod}3$ for $i$ odd. Thus the largest $\mathbb{Z}_2$-submodule of $H^{2i}(\mathbb{Z}_3)\cong\mathbb{Z}_3$ on which $\mathbb{Z}_2$ acts trivially is $\mathbb{Z}_3$ (itself) for $i$ even, and is $0$ for $i$ odd. It now follows that the integral cohomology $H^*(S_3)$ is the same as that which was deduced in Exercise III.10.1, namely, it is $\mathbb{Z}_2$ in the $2\,\text{mod}4$ dimensions and is $\mathbb{Z}_6$ in the $0\,\text{mod}4$ dimensions and is $0$ otherwise (besides the $0^{th}$-dimension in which it is $\mathbb{Z}$).
III.10.1:
The symmetric group $G=S_3$ on three letters is the group of order $3!=6$ whose elements are the permutations of the set $\lbrace 1,2,3\rbrace$. The Sylow $3$-subgroup is generated by the cycle $(1\,2\,3)$, and a Sylow $2$-subgroup is generated by the cycle $(1\,2)$. Noting the semi-direct product representation $S_3=\mathbb{Z}_3\rtimes \mathbb{Z}_2$ where $\mathbb{Z}_2$ acts on $\mathbb{Z}_3$ by conjugation, we have $H^*(S_3)=H^*(S_3)_{(2)}\oplus H^*(S_3)_{(3)}\cong H^*(S_3)_{(2)}\oplus H^*(\mathbb{Z}_3)^{\mathbb{Z}_2}$ by Theorem III.10.3. Now $S_3$ is the unique nonabelian group of order 6, so $D_6\cong S_3$ and we can use Exercise AE.9 which implies that the $\mathbb{Z}_2$-action on $H_{2i-1}(\mathbb{Z}_3)\cong H^{2i}(\mathbb{Z}_3)$ is multiplication by $(-1)^i$. Thus $H^n(\mathbb{Z}_3)^{\mathbb{Z}_2}$ is isomorphic to $\mathbb{Z}_3$ for $n=2i$ where $i$ is even, and is trivial for $n$ odd and $n = 2i$ where $i$ is odd. Taking any Sylow $2$-subgroup $H\cong \mathbb{Z}_2$, Theorem III.10.3 states that $H^*(S_3)_{(2)}$ is isomorphic to the set of $S_3$-invariant elements of $H^*(H)$. In particular we have the monomorphism $H^{2i-1}(S_3)_{(2)}\hookrightarrow H^{2i-1}(H)=0$, so $H^{2i-1}(S_3)_{(2)}=0$. An $S_3$-invariant element $z\in H^{2i}(H)\cong\mathbb{Z}_2$ must satisfy the equation $\text{res}^H_Kz=\text{res}^{gHg^{-1}}_Kgz$, where $K$ denotes $H\cap gHg^{-1}$. If $g\in H$ then $gHg^{-1}=H$ and the above condition is trivially satisfied for all $z$ ($hz=z$ by Proposition III.8.1). If $g\notin H$ then $K=\lbrace 1\rbrace$ because $H$ is not normal in $S_3$ and only contains two elements, so the intersection must only contain the trivial element. But then the image of both restriction maps is zero, so the condition is satisfied for all $z$; thus $H^{2i}(S_3)_{(2)}=\mathbb{Z}_2$. Alternatively, a theorem of Richard Swan states that if $G$ is a finite group such that $\text{Syl}_p(G)$ is abelian and $M$ is a trivial $G$-module, then $\text{Im}(\text{res}^G_{\text{Syl}_p(G)})=H^*(\text{Syl}_p(G),M)^{N_G(\text{Syl}_p(G))}$. It is a fact that $N_{S_3}(\mathbb{Z}_2)=\mathbb{Z}_2$, so taking $G=S_3$ and $H=\text{Syl}_2(S_3)\cong\mathbb{Z}_2$ and $M=\mathbb{Z}$ we have $\text{Im}(\text{res}^{S_3}_H)=(\mathbb{Z}_2)^{\mathbb{Z}_2}=\mathbb{Z}_2$ in the even-dimensional case. Since any invariant is in the image of the above restriction map (by Theorem III.10.3), the result $H^{2i}(S_3)_{(2)}=\mathbb{Z}_2$ follows.
Recapping: $H^n(S_3)\cong$
$\mathbb{Z}$ for $n=0$,
$\mathbb{Z}_6$ for $n\equiv 0\;\text{mod}\,4$ with $n\ne 0$,
$\mathbb{Z}_2$ for $n\equiv 2\;\text{mod}\,4$,
$0$ otherwise
