Write the integers from 1 to 60 one after another to make the single large number N =…

(a) $10 \equiv 1 \pmod 9$, so $N$ leaves the same remainder as its digit sum. The digit sum is 426, and 426 leaves remainder $3$ on division by $9$.

(b) $10 \equiv -1 \pmod{11}$, so $N$ is congruent to its digits added with alternating signs, starting with $+$ at the units digit. The number has 111 digits. The alternating sum is $74$, and $74 \equiv 8 \pmod{11}$.

(c) $10\,000 = 16 \times 625$, so $N$ and its last four digits $5960$ leave the same remainder modulo $16$: $5960 = 16 \times 372 + 8$.

(d) $100 \equiv 1 \pmod{99}$, so $N$ is congruent to the sum of its two-digit blocks counted from the right (the leftmost block has one digit). There are 56 blocks and their sum is 2010, and $2010 \equiv 30 \pmod{99}$. The answers for $9$ and $11$ agree: $30$ is the only remainder below $99$ that is $3$ modulo $9$ and $8$ modulo $11$.