Answer: A \(90\%\) confidence interval for \(\mu_X-\mu_Y\) is \(\boxed{[-0.506,-0.084]}\), and the possible values are \(\boxed{\alpha<2.2}\).

(i) The sample means are

\[\bar{x}=\frac{45.5}{50}=0.91,\qquad \bar{y}=\frac{72.3}{60}=1.205.\]

The unbiased estimates of the population variances are

\[s_X^2=\frac{19.56}{49}=0.3992,\]

and

\[s_Y^2=\frac{30.25}{59}=0.5127.\]

For the difference \(\mu_X-\mu_Y\), the standard error is

\[\sqrt{\frac{s_X^2}{50}+\frac{s_Y^2}{60}}.\]

So

\[\mathrm{SE}=\sqrt{\frac{0.3992}{50}+\frac{0.5127}{60}}=0.1286.\]

The observed difference is

\[\bar{x}-\bar{y}=0.91-1.205=-0.295.\]

For a \(90\%\) confidence interval,

\[z_{0.95}=1.645.\]

So the margin of error is

\[1.645(0.1286)=0.211.\]

Therefore the confidence interval for \(\mu_X-\mu_Y\) is

\[-0.295\pm0.211.\]

Hence

\[\boxed{[-0.506,-0.084]}.\]

Equivalently, for \(\mu_Y-\mu_X\), the interval would be \([0.084,0.506]\).

(ii) The test is two-tailed because it asks whether there is any difference between the two means.

The test statistic is

\[z=\frac{1.205-0.91}{0.1286}=2.29.\]

From the standard normal distribution,

\[\Phi(2.29)\approx0.989.\]

So the two-tailed \(p\)-value is

\[2(1-0.989)=0.022.\]

As a percentage, this is about

\[2.2\%.\]

The test does not provide evidence of a difference when the significance level is less than this limiting value.

Therefore

\[\boxed{\alpha<2.2}.\]