= Solution
Write the common variance as $\sigma^2$. Independence and the <normal distribution> give
$$
\boxed{\bar X\sim N(\mu_X,\sigma^2/m),\qquad S_{XX}/\sigma^2\sim\chi^2_{m-1}.}
$$
Moreover $\bar X$ and $S_{XX}$ are independent. To see both assertions, standardize the observations into an isotropic Gaussian vector and choose an orthonormal basis with first vector $(1,\ldots,1)/\sqrt m$. Its first coordinate is $\sqrt m(\bar X-\mu_X)/\sigma$; the remaining independent standard normal coordinates have squared sum $S_{XX}/\sigma^2$. The same argument applies to the $Y$ sample, independently of the first sample.
Under the null hypothesis, the difference of sample means is $N(0,\sigma^2(1/m+1/n))$. The residual sums of squares add to an independent <chi-squared distribution> with $\nu=m+n-2$ degrees of freedom. Define the <pooled sample variance> and test statistic by
$$
S_p^2=\frac{S_{XX}+S_{YY}}{m+n-2},\qquad
\boxed{T=\frac{\bar X-\bar Y}{S_p\sqrt{1/m+1/n}}.}
$$
For $\nu>0$, under $H_0$ this is $Z/\sqrt{U/\nu}$ with independent $Z\sim N(0,1)$ and $U\sim\chi^2_\nu$. Hence \b[$T\sim t_{m+n-2}$ under the null], the <Student t-distribution>.
Let $t_{\nu,0.99}$ be its 99th percentile. For the specified one-sided alternative, reject $H_0$ precisely when
$$
\boxed{T>t_{m+n-2,0.99}.}
$$
Because this distribution is continuous, the rejection probability under $H_0$ is exactly $0.01$, irrespective of the nuisance mean and variance. Large positive values supply evidence for $\mu_X>\mu_Y$; a two-sided or 0.995 cutoff would test a different alternative. The common-variance and independence assumptions are essential to this exact <Student t-test>. If both samples have size one, there are no residual degrees of freedom and the proposed unknown-variance test cannot be formed.
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