= Solution
For finite <sets> $A_1,\ldots,A_s$, the <inclusion-exclusion principle> states
$$
\left|\bigcup_{i=1}^sA_i\right|
=\sum_{\varnothing\ne J\subseteq\{1,\ldots,s\}}(-1)^{|J|+1}
\left|\bigcap_{j\in J}A_j\right|.
$$
To prove it, count the contribution of an element lying in exactly $r\ge1$ of the <sets>. Its total coefficient is
$$
\sum_{j=1}^r(-1)^{j+1}\binom rj=1-(1-1)^r=1,
$$
by the <binomial theorem>. An element in none contributes zero. Summing these elementwise counts proves the formula. Applying the same argument to <indicator functions> and taking <expectations> proves the corresponding probability formula.
Factor $4199=13\cdot17\cdot19$. An integer is <coprime> to this product precisely when none of its three <prime factors> divides it. <Inclusion-exclusion> applied to the multiples of those <prime numbers> among $1,\ldots,4199$ gives <Euler's totient function>
$$
\begin{aligned}
\varphi(4199)
&=4199-(323+247+221)+(19+17+13)-1\\
&=4199\left(1-\frac1{13}\right)\left(1-\frac1{17}\right)\left(1-\frac1{19}\right)\\
&=12\cdot16\cdot18=\boxed{3456}.
\end{aligned}
$$
For the survey, let $H,D,S$ be the three detestation <events> and use proportions as <probabilities>. The union $H\cup D$ is the reported total union with the only-$S$ group removed, so $\mathbb P(H\cup D)=0.90-0.27=0.63$. The two-set <inclusion-exclusion principle> gives $\mathbb P(H\cap D)=0.45+0.28-0.63=0.10$. Removing the all-three group leaves
$$
\boxed{\mathbb P((H\cap D)\setminus S)=0.10-0.06=0.04.}
$$
The required proportion is \b[4%]. The separately reported total $S$ proportion is consistent with, but unnecessary for, this calculation.
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