= Solution
A nondegenerate <distribution function> $G$ is a <max-stable distribution> if, for every integer $r\geq1$, there are $a_r>0$ and $b_r\in\mathbb R$ such that
$$
G(a_rx+b_r)^r=G(x)\qquad(x\in\mathbb R).
$$
Equivalently, the suitably centered and positively scaled <sample maximum> of $r$ <independent and identically distributed random variables> with law $G$ again has law $G$. The <extremal types theorem> states that every nondegenerate limiting law of normalized <sample maxima> is a <max-stable distribution>. Thus this property identifies all possible limiting types, although a particular starting <distribution function> need not admit any such limiting normalization.
The three standard types are the <Gumbel distribution>
$$
\Lambda(x)=\exp(-e^{-x}),\qquad x\in\mathbb R,
$$
the <Fréchet distribution>, with $\alpha>0$,
$$
\Phi_\alpha(x)=\begin{cases}0,&x\leq0,\\ \exp(-x^{-\alpha}),&x>0,\end{cases}
$$
and the <negative Weibull distribution>, with $\alpha>0$,
$$
\Psi_\alpha(x)=\begin{cases}\exp(-(-x)^\alpha),&x<0,\\1,&x\geq0.\end{cases}
$$
They are <max-stable distributions>, as follows respectively by taking $(a_r,b_r)=(1,\log r)$, $(r^{1/\alpha},0)$, and $(r^{-1/\alpha},0)$. The necessary and sufficient classification is: \b[$G$ is max-stable exactly when $G(x)=H((x-b)/a)$ for $a>0$, $b\in\mathbb R$, and one of these three types $H$]. A positive <affine map> sends a <max-stable distribution> to another <max-stable distribution>.
For the requested threshold equivalence, set $p_n=1-F(u_n)$. By <independence>,
$$
\mathbb P(X_{(n)}\leq u_n)=(1-p_n)^n.
$$
If $np_n\to\tau<\infty$, then $p_n\to0$ and the expansion $\log(1-p)=-p+O(p^2)$ gives $n\log(1-p_n)\to-\tau$, since $np_n^2=(np_n)p_n\to0$. Exponentiation proves the claimed limit. Conversely, if $(1-p_n)^n\to e^{-\tau}>0$, then $p_n\to0$: any subsequence with $p_n\geq\delta>0$ would have its powers tending to zero. Taking <logarithms> now gives $-n\log(1-p_n)\to\tau$. Since $-\log(1-p)/p\to1$, including the continuous extension at $p=0$, it follows that $np_n\to\tau$. Hence
$$
\boxed{\mathbb P(X_{(n)}\leq u_n)\to e^{-\tau}\ \Longleftrightarrow\ n\{1-F(u_n)\}\to\tau.}
$$
The exceedance count $S_n$ has the <binomial distribution> with parameters $n,p_n$. For $\tau>0$ and each fixed nonnegative integer $s$, its <probability mass function> satisfies
$$
\mathbb P(S_n=s)=\frac{(n)_s}{s!}p_n^s(1-p_n)^{n-s}\longrightarrow e^{-\tau}\frac{\tau^s}{s!},
$$
where $(n)_s=n(n-1)\cdots(n-s+1)$ and $(n)_0=1$. Indeed $(n)_s/n^s\to1$, $(np_n)^s\to\tau^s$, and $(1-p_n)^{n-s}\to e^{-\tau}$. If $\tau=0$, the bound $\mathbb P(S_n\geq1)\leq\mathbb ES_n=np_n\to0$ gives the degenerate <Poisson distribution> at zero. Summing the finitely many terms proves, in both cases,
$$
\boxed{\mathbb P(S_n\leq k)\longrightarrow e^{-\tau}\sum_{s=0}^k\frac{\tau^s}{s!},\qquad k=0,1,2,\ldots.}
$$
This derives the needed instance of the <Poisson limit theorem> directly.
Finally fix $x$ with $G(x)>0$ and put $u_n=a_nx+b_n$. The assumed limit for the normalized <sample maximum> and the proved equivalence give $np_n\to-\log G(x)$. The second-largest <order statistic> is at most $u_n$ exactly when at most one observation is strictly greater than $u_n$. This event identity holds even if observations have ties. Apply the preceding <Poisson limit theorem> with $k=1$ to obtain
$$
\boxed{\mathbb P\!\left(\frac{W_n-b_n}{a_n}\leq x\right)\longrightarrow G(x)\{1-\log G(x)\}.}
$$
The endpoint $G(x)=1$ corresponds to $\tau=0$ and is already included. This is the case $r=2$ of the <exceedance count limit for extreme order statistics>.
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