Answer: \(E_6=\boxed{6.35}\), \(E_{\ge7}=\boxed{4.77}\), and the Poisson model is suitable at the \(5\%\) level.

(i) The mean is

\[\bar{x}=\frac{\sum xf}{100}.\]

Using the table,

\[\sum xf=0(4)+1(9)+2(18)+3(26)+4(20)+5(16)+6(7)=325.\]

Therefore

\[\bar{x}=\frac{325}{100}=3.25,\]

as required.

(ii) For a Poisson distribution with mean \(\lambda=3.25\),

\[P(X=r)=\frac{e^{-3.25}3.25^r}{r!}.\]

Since there are 100 nights, the expected frequency for \(X=r\) is

\[100\frac{e^{-3.25}3.25^r}{r!}.\]

For \(x=3\),

\[E_3=100\frac{e^{-3.25}3.25^3}{3!}=22.18,\]

which is the given value.

For \(x=6\),

\[E_6=100\frac{e^{-3.25}3.25^6}{6!}=\boxed{6.35}.\]

For \(x\ge7\), use the total 100:

\[E_{\ge7}=100-(E_0+E_1+E_2+E_3+E_4+E_5+E_6).\]

Using the expected frequencies,

\[E_{\ge7}=100-(3.88+12.60+20.48+22.18+18.02+11.72+6.35).\]

Hence

\[E_{\ge7}=\boxed{4.77}.\]

(iii) We test whether the Poisson distribution is a suitable model.

Because some expected frequencies are below 5, combine the first two cells and the last two cells:

\[\begin{array}{c|cccccc} \text{cell}&0,1&2&3&4&5&6,\ge7\\ \hline O&13&18&26&20&16&7\\ E&16.48&20.48&22.18&18.02&11.72&11.12 \end{array}\]

The test statistic is

\[\chi^2=\sum\frac{(O-E)^2}{E}.\]

So

\[\chi^2=0.735+0.300+0.658+0.218+1.563+1.527=5.00.\]

There are 6 grouped cells. Since one parameter \(\lambda\) has been estimated from the data, the degrees of freedom are

\[6-1-1=4.\]

At the \(5\%\) significance level, the critical value is

\[\chi^2_{4,0.95}=9.488.\]

Since

\[5.00<9.488,\]

we do not reject the Poisson model. Therefore the Poisson distribution is a suitable model for the data at the \(5\%\) significance level.