= Solution
Put $m=2n$ and let $\alpha$ be the identity class of $S^m$. With the product-cell orientation used in part (i), form the folding map $\nabla:S^m\vee S^m\to S^m$, which is identity on each summand. The top cell of $S^m\times S^m$ is attached by the product of the two inclusions, so folding gives the attaching map $[\alpha,\alpha]_W$. Thus there is a <cellular map>
$$
F:S^m\times S^m\longrightarrow C,
\qquad C=S^m\cup_{[\alpha,\alpha]_W}D^{2m},
$$
which is the fold on the lower skeleton and degree one on the top relative cell.
Let $a,b\in H^m(S^m\times S^m;\mathbb Z)$ be the classes from the two factors, and let $x\in H^m(C;\mathbb Z)$ restrict to the generator of the bottom sphere. Orient the top class $z\in H^{2m}(C;\mathbb Z)$ so that $F^*z=ab$. Then $F^*x=a+b$. Since $a^2=b^2=0$ and $m$ is even, <graded commutative algebra> gives
$$
F^*(x^2)=(a+b)^2=ab+ba=2ab=F^*(2z).
$$
The map $F^*$ is an isomorphism in degree $2m$, because of its degree-one top-cell map. Therefore $x^2=2z$. By the definition of the <Hopf invariant> this proves
$$
\boxed{H([\alpha,\alpha])=2.}
$$
The shifted Lie convention in part (i) multiplies the geometric bracket by $(-1)^m=1$, so it gives the same square and the same invariant. Reversing the orientation chosen for the attached top cell would reverse both its generator and the numerical sign convention; with the standard orientation the value is the stated positive two.
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