= Solution
Assume the usual continuous nonnegative valuation distribution, so ties occur only on <zero-probability events>. Put $t=F(v)$. In a monotone symmetric <Bayesian Nash equilibrium>, a type $v$ is first with probability $t^{n-1}$ and second with probability $(n-1)(1-t)t^{n-2}$. Its <rank-order expected prize allocation> in $G_1$ is therefore
$$
a_1(v)=2t^{n-1}+(n-1)(1-t)t^{n-2}.
$$
The <all-pay effort identity> gives $b_1(v)=\int_{\underline v}^v s a_1'(s)ds$. It also verifies equilibrium directly: a type $v$ imitating type $z$ has utility $v a_1(z)-b_1(z)$, whose derivative is $(v-z)a_1'(z)$, so the true type is a <best response>.
In the first version of $G_2$, the two contests have expected allocations
$$
a_{21}(v)=t^{n-1},\qquad
a_{22}(v)=t^{n-1}+(n-1)(1-t)t^{n-2}.
$$
There is no common effort budget, and <quasilinear utility> makes the two effort choices separable. Since $a_{21}+a_{22}=a_1$, adding their <all-pay effort identities> yields
$$
\boxed{b_{21}(v)+b_{22}(v)=b_1(v),\qquad
\mathbb E[\text{total effort in }G_2]=\mathbb E[\text{total effort in }G_1].}
$$
The equality holds type by type for aggregate effort, rather than only after taking expectations. The within-player correlation of the two efforts does not enter these additive expected payoffs.
For the second version of $G_2$, let $V_{[1]}\geq\cdots\geq V_{[n]}$ denote <descending order statistics>. The <expected effort in a rank-order contest> with prize vector $(2,1,0,\ldots)$ is
$$
\mathbb E E_1=\mathbb E V_{[2]}+2\mathbb E V_{[3]}.
$$
Equivalently, decompose the allocation into a unit award to the best player and a unit award to each of the best two players, then use <revenue equivalence>: the corresponding total auction payments are $V_{[2]}$ and $2V_{[3]}$. Two separate first-place contests with prize values one and two instead generate
$$
\mathbb E E_2=3\mathbb E V_{[2]}.
$$
Consequently
$$
\boxed{\mathbb E E_2-\mathbb E E_1
=2\mathbb E[V_{[2]}-V_{[3]}]\geq0.}
$$
For a nondegenerate continuous distribution, the inequality is strict. No regularity of <virtual valuations> is needed for this comparison. With a <uniform distribution> on $[0,1]$, the two totals are $(3n-5)/(n+1)$ and $3(n-1)/(n+1)$, giving a difference of $4/(n+1)$.
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