1. The two reagents given in the options are:

   • $\mathrm{CH_3OH}/\mathrm{H^+},\ \Delta$ : this is used for esterification of a carboxylic acid to form a methyl ester.
   • $\left(\mathrm{CH_3CO}\right)_2\mathrm{O}/\mathrm{H^+}$ : this is used for acetylation of an $-OH$ group, i.e. conversion of an alcohol/phenol into an acetate ester.

2. In such standard organic conversion questions involving compounds having both $-COOH$ and $-OH$ functionality, the order of reagents matters:

   • If we first use acetic anhydride, the hydroxyl group gets acetylated.
   • Then treatment with $\mathrm{CH_3OH}/\mathrm{H^+},\ \Delta$ converts the carboxylic acid group into methyl ester.

3. Let us examine the options:

   A: $\mathrm{CH_3OH}/\mathrm{H^+},\Delta$ first, then acetic anhydride
   This would first esterify the acid group, and then acetylate the hydroxyl group. This order is possible in some cases, but for the standard protected-product sequence asked here, the intended order is generally acetylation first and esterification second.

   B: $\left(\mathrm{CH_3CO}\right)_2\mathrm{O}/\mathrm{H^+}$ first, then $\mathrm{CH_3OH}/\mathrm{H^+},\Delta$
   First acetylation of $-OH$, then esterification of $-COOH$ to methyl ester. This matches the expected transformation.

   C: $\mathrm{CH_3OH}/\mathrm{H^+},\Delta$ and again $\mathrm{CH_3OH}/\mathrm{H^+},\Delta$
   Same reagent twice cannot account for both acetylation and methyl ester formation.

   D: $\left(\mathrm{CH_3CO}\right)_2\mathrm{O}/\mathrm{H^+}$ and again $\left(\mathrm{CH_3CO}\right)_2\mathrm{O}/\mathrm{H^+}$
   Same reagent twice cannot account for both transformations either.

4. Therefore, the correct pair of reagents is: $X = \left(\mathrm{CH_3CO}\right)_2\mathrm{O}/\mathrm{H^+}, \qquad Y = \mathrm{CH_3OH}/\mathrm{H^+},\Delta$

Hence, the correct option is B.