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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 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 $$H^d(BSO_{\infty};\mathbb{Z})\cong 0,\quad (d=1,2)$$ $$H^4(BSO_{\infty};\mathbb{Z})\cong \mathbb{Z},$$ $$H^5(BSO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2.$$

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 $$H^1(BO_{\infty};\mathbb{Z})\cong 0,$$ $$H^i(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2, \quad (i=2,3)$$ $$H^4(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}\oplus (\mathbb{Z}/2)^2,$$ $$H^5(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2\oplus \mathbb{Z}/2.$$

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 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 $$H^d(BSO_{\infty};\mathbb{Z})\cong 0,\quad (d=1,2),$$ $$H^d(BSO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2 ,\quad (d=3,5), $$ $$H^4(BSO_{\infty};\mathbb{Z})\cong \mathbb{Z}.$$

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

All of these lead to $$H^1(BO_{\infty};\mathbb{Z})\cong 0,$$ $$H^i(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2, \quad (i=2,3)$$ $$H^4(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}\oplus (\mathbb{Z}/2)^2,$$ $$H^5(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2\oplus \mathbb{Z}/2.$$

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 $$H^5(BO_{\infty};Z)\cong Z/2\oplus Z/2,$$

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 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 $$H^d(BSO_{\infty};\mathbb{Z})\cong 0,\quad (d=1,2)$$ $$H^4(BSO_{\infty};\mathbb{Z})\cong \mathbb{Z},$$ $$H^5(BSO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2.$$

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 $$H^1(BO_{\infty};\mathbb{Z})\cong 0,$$ $$H^i(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2, \quad (i=2,3)$$ $$H^4(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}\oplus (\mathbb{Z}/2)^2,$$ $$H^5(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2\oplus \mathbb{Z}/2.$$

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

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 $$H^5(BO_{\infty};Z)\cong Z/2\oplus Z/2,$$