The Theorem 1.5 and 1.6 you quote give the answer.

More precisely, for $SO$, in the range $d<6$, the only polynomial generators
are $p_1$ which has the degree 4, $\delta(w_2)$ with degree 3 and $\delta(w_4)$ with degree 5. The only  relations are $2\delta(w_{2i})=0$, which gives $$H^d(BSO_{\infty};Z)\cong 0$$ if $d=1,2$
 $$H^d(BSO_{\infty};Z)\cong Z/2,$$ if $d=5$
and $$H^4(BSO_{\infty};Z)\cong Z.$$

In the case of $BO$, there are more generators $\delta(w_1)$, $\delta(w_2)$
 $\delta(w_1w_2)$ and$\delta(w_4)$  in degrees 2,3,4 and 5.  Thus in degrees 4 and  5 we also have  products $\delta(w_1)^2$ $\delta(w_1)\delta(w_2)$.

All of these lead to
$$H^1(BO_{\infty};Z)\cong 0$$

$$H^4(BO_{\infty};Z)\cong Z\oplus (Z/2)^2,$$

$$H^i(BO_{\infty};Z)\cong Z/2.$$
for $i=2,3$ and
$$H^5(BO_{\infty};Z)\cong Z/2\oplus Z/2,$$