The Theorem 1.5 and 1.6 you quote give the answer.
More precisely, for $SO$, in the range $d<8$, the only polynomial generators are $p_1$ which has the degree 4 and $\delta(w_2)$ with degree 3. The only relation is $2\delta(w_2)=0$, which gives $$H^d(BSO_{\infty};Z)\cong 0$$ if $d=1,2,5$ $$H^3(BSO_{\infty};Z)\cong Z/2,$$ and $$H^4(BSO_{\infty};Z)\cong Z.$$
In the case of $BO$, there are more generators $\delta(w_1)$, $\delta(w_2)$ and $\delta(w_1w_2)$ in degrees 2,3, and 4. Thus in degree 5 we also have a product $\delta(w_1)\delta(w_2)$.
All of these lead to $$H^1(BO_{\infty};Z)\cong 0$$
$$H^4(BO_{\infty};Z)\cong Z\oplus Z/2,$$ and $$H^i(BO_{\infty};Z)\cong Z/2.$$ for $i=2,3 \mbox{ and }5$