= Solution
The space is the <mapping torus> of the <antipodal map> $A:S^2\to S^2$. Its <mapping degree> is $(-1)^3=-1$, so transporting a fiber orientation once around the base circle reverses it. \b[$X$ is nonorientable.]
On integral fiber <homology>, $A_*$ is the identity in degree zero and minus the identity in degree two. The <Wang sequence>, whose relevant map is $1-A_*$, therefore gives
$$
H_0(X;\mathbb Z)=H_1(X;\mathbb Z)=\mathbb Z,\qquad H_2(X;\mathbb Z)=\mathbb Z/2,\qquad H_3(X;\mathbb Z)=0.
$$
The <universal coefficient theorem for cohomology> shifts the torsion contribution into degree three, because $\operatorname{Ext}_{\mathbb Z}(\mathbb Z/2,\mathbb Z)=\mathbb Z/2$. Hence
$$
\boxed{H^q(X;\mathbb Z)=\begin{cases}\mathbb Z,&q=0,1,\\\mathbb Z/2,&q=3,\\0,&\text{otherwise}.\end{cases}}
$$
Let $t\in H^1(X;\mathbb Z)$ be pulled back from the base circle, and $u$ the nonzero degree-three class. The square of the circle class vanishes, so $t^2=0$. All other products of positive-degree classes land above dimension three and vanish. This gives the <cohomology ring of the antipodal two-sphere mapping torus>
$$
\boxed{H^*(X;\mathbb Z)=\mathbb Z[t,u]/(t^2,tu,u^2,2u),\qquad |t|=1,\ |u|=3.}
$$
Modulo two, $A^*$ is the identity on fiber cohomology. The cohomological <Wang sequence> shows that restriction $H^2(X;\mathbb F_2)\to H^2(S^2;\mathbb F_2)$ is onto; choose $y$ mapping to its generator. The classes $1,y$ restrict to a fiber basis, so the <Leray-Hirsch theorem> makes the total cohomology free over $H^*(S^1;\mathbb F_2)$ on those two classes. Write $\bar t$ for the mod-two base class. The basis is $1,\bar t,y,\bar t y$, and in particular $\bar t y\ne0$. Since $\bar t^2=0$ by pullback and $y^2=0$ by dimension, the <mod-two cohomology ring of the antipodal two-sphere mapping torus> is
$$
\boxed{H^*(X;\mathbb F_2)=\mathbb F_2[\bar t,y]/(\bar t^2,y^2),\qquad |\bar t|=1,\ |y|=2.}
$$
The argument of Question 3 works over any field, so its use here over $\mathbb F_2$ is justified. The ring agrees with that of $S^1\times S^2$; a cohomology ring alone does not determine orientability.
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