Answer: \(\boxed{p=0.4}\), \(\boxed{P(X=5)=0.0518}\), and \(\boxed{P(3\leq X\leq7)=0.332}\).

The random variable \(X\) is the number of attempts required to get the first pass, so \(X\) has 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 \(3.75\), so

\[\frac{1-p}{p^2}=3.75.\]

Since \(3.75=\frac{15}{4}\), multiply by \(4p^2\):

\[4(1-p)=15p^2.\]

Therefore

\[15p^2+4p-4=0,\]

as required.

Now solve the quadratic:

\[15p^2+4p-4=(5p-2)(3p+2)=0.\]

So

\[p=\frac25\quad\text{or}\quad p=-\frac23.\]

A probability cannot be negative, so

\[\boxed{p=\frac25=0.4}.\]

(ii) To have \(X=5\), the driver fails the first four attempts and passes on the fifth. Hence

\[P(X=5)=(1-p)^4p.\]

Using \(p=0.4\),

\[P(X=5)=0.6^4(0.4)=0.05184.\]

So

\[\boxed{P(X=5)=0.0518}\]

to 3 significant figures.

(iii) The event \(3\leq X\leq7\) means the first success happens on attempts \(3,4,5,6\) or \(7\). A quick way is

\[P(3\leq X\leq7)=P(X\geq3)-P(X\geq8).\]

Now \(P(X\geq3)\) means the first two attempts are failures, and \(P(X\geq8)\) means the first seven attempts are failures. Therefore

\[P(3\leq X\leq7)=0.6^2-0.6^7.\]

So

\[P(3\leq X\leq7)=0.36-0.0279936=0.3320064.\]

Hence

\[\boxed{P(3\leq X\leq7)=0.332}.\]