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 $i=2,3 \mbox{ and }5$