Step 1: Recall what makes an oxime.
Hydroxylamine $NH_2OH$ reacts with a free carbonyl group such as the aldehyde $-CHO$ of glucose to give an oxime. So a free aldehyde must be available for an oxime to form.
Step 2: Look at reaction I.
In reaction I, plain D-glucose meets $NH_2OH$. Glucose has a free $-CHO$ group (in its open chain form), so it readily forms glucose oxime.
Step 3: Conclude for reaction I.
So an oxime IS formed in reaction I.
Step 4: Look at reaction II.
Here glucose is first treated with acetic anhydride $(CH_3CO)_2O$. This acetylates the hydroxyl groups and locks glucose in its cyclic acetylated form, so the aldehyde group is no longer free.
Step 5: Apply the oxime rule to reaction II.
With the carbonyl now blocked, $NH_2OH$ has nothing to react with, so NO oxime forms in reaction II.
Step 6: Combine the two results.
Oxime forms in reaction I but not in reaction II.
\[ \boxed{\text{Oxime is formed in reaction I but oxime is not formed in reaction II}} \]