A shop accepts payment in coins of two kinds only, worth 13 and 17 rupees…

(a) For coprime $a, b$ the greatest amount not payable is $ab - a - b = 221 - 13 - 17 = 191$.

(b) Exactly half of the integers $0, 1, \dots, ab - a - b$ are payable, which gives $\dfrac{(a - 1)(b - 1)}{2} = \dfrac{12 \times 16}{2} = 96$ amounts that are not payable.

(c) Two ways $(x, y)$ and $(x', y')$ for one amount need $13(x - x') = 17(y' - y)$, so $17$ divides $x - x'$ and $13$ divides $y' - y$. The smallest change is $x$ up by $17$ and $y$ down by $13$, which needs at least $13 \times 17 = 221$ rupees. The amount $221$ itself is paid as $17$ coins of $13$ or as $13$ coins of $17$. So the least is $221$.

(d) Use as many $17$ rupee coins as possible. Since $1000 - 17y$ must be a multiple of $13$, take $y = 55$: then $1000 - 17 \times 55 = 5 \times 13$, a total of $60$ coins. Using fewer $17$ rupee coins needs more coins, because a $17$ rupee coin does more work than a $13$ rupee coin.

(e) From $13x \equiv 2026 \pmod{17}$ and $13 \times 4 \equiv 1 \pmod{17}$, we get $x \equiv 4 \times 2026 \equiv 12 \pmod{17}$. The values of $x$ are $12, 12 + 17, \dots$ up to at most $2026 \div 13 = 155.8$. The last one is $148$, so there are $9$ ways.