Answer: \(\boxed{p=\dfrac34}\), \(\boxed{\mathrm P(X=3)=\dfrac{3}{64}}\), \(\boxed{\mathrm P(X\geq4)=\dfrac{1}{64}}\), and \(\boxed{N=5}\).

The random variable \(X\) is the day on which the first snowfall occurs. This is a geometric distribution with parameter \(p\).

(i) For this form of the geometric distribution,

\[\operatorname{Var}(X)=\frac{1-p}{p^2}.\]

We are told that the variance is \(\frac49\), so

\[\frac{1-p}{p^2}=\frac49.\]

Multiply by \(9p^2\):

\[9(1-p)=4p^2.\]

Therefore

\[4p^2+9p-9=0,\]

as required.

Now factorise:

\[4p^2+9p-9=(4p-3)(p+3).\]

So

\[p=\frac34\quad\text{or}\quad p=-3.\]

A probability cannot be negative, so

\[\boxed{p=\frac34}.\]

(ii) The first snowfall is on 3 November when there is no snow on the first two days and then snow on the third day. Thus

\[\mathrm P(X=3)=(1-p)^2p.\]

With \(p=\frac34\),

\[\mathrm P(X=3)=\left(\frac14\right)^2\left(\frac34\right)=\boxed{\frac{3}{64}}.\]

(iii) The first snowfall is not before 4 November when there is no snow on the first three days. Therefore

\[\mathrm P(X\geq4)=(1-p)^3=\left(\frac14\right)^3=\boxed{\frac{1}{64}}.\]

(iv) We need

\[\mathrm P(X\leq N)>0.999.\]

For the geometric distribution,

\[\mathrm P(X\leq N)=1-(1-p)^N.\]

So

\[1-\left(\frac14\right)^N>0.999.\]

This means

\[\left(\frac14\right)^N<0.001.\]

Take logarithms:

\[N>\frac{\log(0.001)}{\log(0.25)}=4.98\ldots.\]

Therefore the least integer value is

\[\boxed{N=5}.\]