Step 1: Sort the question.
There is one term to define and two pairs to compare. A short, clear comparison for each pair is enough.
Step 2: Part (a), reducing sugar.
Think of what the test does. Fehling's solution has $(\$mathrm{Cu^{2+}}$) and Tollens' reagent has \$($mathrm{Ag^+}\$). A sugar that turns these into red $(\$mathrm{Cu_2O}$) or a silver mirror is a reducing sugar.
It can do this because it has a free \$($mathrm{-CHO}\$) group, or a ring form that can open to give one.
Example: glucose. Non-example: sucrose.
Step 3: Part (b)(i), compare the two protein types.
Shape: fibrous proteins are long fibres or sheets; globular proteins are coiled into balls.
Water: fibrous proteins do not dissolve; globular proteins dissolve.
Role and example: fibrous proteins give structure (hair keratin, silk fibroin); globular proteins do work such as enzymes and hormones (insulin, haemoglobin).
Structure: fibrous proteins mainly have secondary structure; globular proteins also have tertiary and quaternary folding.
Step 4: Part (b)(ii), compare the two units.
Start from the base. Base + sugar (ribose or deoxyribose) joined at C-1 gives a nucleoside.
Add a phosphate ester at C-5 of the sugar and you get a nucleotide.
So the only difference is the phosphate group.
Examples: adenosine is a nucleoside; adenosine monophosphate (AMP) is a nucleotide.
Final Answer:
A reducing sugar reduces Fehling's and Tollens' reagents because of a free carbonyl group. Fibrous proteins are insoluble fibres; globular proteins are soluble spheres. A nucleoside lacks the phosphate that a nucleotide has. $[ \$boxed{$text{Nucleoside} + \$text{phosphate} = $text{Nucleotide}} \$]