Answer: \(\boxed{a=78.49}\), \(\boxed{b=11.41}\), and the manager's belief is not justified at the \(1\%\) significance level.

(i) The sample mean is

\[\bar{x}=\frac{\sum xf}{250}.\]

Calculate the numerator:

\[0(48)+1(69)+2(78)+3(32)+4(22)+5(1)+6(0)=414.\]

Hence

\[\bar{x}=\frac{414}{250}=1.656,\]

as required.

(ii) If the number of damaged pots has a binomial distribution with \(n=6\), then

\[E(X)=6p.\]

Using the sample mean as the estimate of \(E(X)\),

\[6p=1.656.\]

So

\[p=0.276,\qquad q=1-p=0.724.\]

The expected frequency for \(x=2\) is

\[a=250\binom62q^4p^2.\]

Thus

\[a=250\binom62(0.724)^4(0.276)^2=\boxed{78.49}.\]

The expected frequency for \(x=4\) is

\[b=250\binom64q^2p^4.\]

Therefore

\[b=250\binom64(0.724)^2(0.276)^4=\boxed{11.41}.\]

(iii) Test

\[H_0:\text{the binomial distribution is a suitable model}.\]

The expected frequencies for \(x=5\) and \(x=6\) are less than 5, so combine the classes \(x=4,5,6\).

The observed and expected frequencies used are

\[\begin{array}{c|ccccc} x&0&1&2&3&4,5,6\\ \hline O&48&69&78&32&23\\ E&36.01&82.36&78.49&39.89&13.26 \end{array}\]

Now compute

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

So

\[\chi^2=\frac{(48-36.01)^2}{36.01}+\frac{(69-82.36)^2}{82.36}+\frac{(78-78.49)^2}{78.49}+\frac{(32-39.89)^2}{39.89}+\frac{(23-13.26)^2}{13.26}.\]

This gives

\[\chi^2=14.9\text{ approximately}.\]

There are \(5\) combined cells and \(1\) estimated parameter, so the degrees of freedom are

\[5-1-1=3.\]

At the \(1\%\) significance level,

\[\chi^2_{3,0.99}=11.34.\]

Since

\[14.9>11.34,\]

we reject \(H_0\).

Therefore the binomial distribution does not fit the data well enough at the \(1\%\) level, so the manager's belief is not justified.