1.
(i) Solution: From the value column in Table 1, we know that the largest number not exceeding 420 is 256, and $2^{8}=256$. From $420-256=164$, and from the value column in Table 1, we know that the largest number not exceeding 164 is 128, and $2^{7}=128$. From $164-128=36$, and from the value column in Table 1, we know that the largest number not exceeding 36 is 32, and $2^{5}=32$. From $36-32=4$, and from the value column in Table 1, we know that the largest number not exceeding 4 is 4, and $2^{2}=4$. Since
$$\begin{aligned} 420 & =256+128+32+4 \\ & =1 \times 2^{8}+1 \times 2^{7}+0 \times 2^{5}+1 \times 2^{3} \\ & +0 \times 2^{4}+0 \times 2^{3}+1 \times 2^{2}+0 \times 2+0 \end{aligned}$$

Therefore, $420=(110100100)_{2}$
(ii) Solution: From the value column in Table 1, we know that the largest number not exceeding 2640 is 2048, and $2^{\mathrm{n}}=2048$. From $2640-2048=592$, and from the value column in Table 1, we know that the largest number not exceeding 592 is 512, and $2^{9}=512$. From $592-512=80$, and from the value column in Table 1, we know that the largest number not exceeding 80 is 64, and $2^{5}=64$. From $80-64=16$, and from the value column in Table 1, we know that the largest number not exceeding 16 is 16, and $2^{4}=16$. Since
$$\begin{aligned} 2640= & 2048+512+64+16=1 \times 2^{11} \\ & +0 \times 2^{10}+1 \times 2^{9}+0 \times 2^{8}+0 \times 2^{7} \\ & +1 \times 2^{6}+0 \times 2^{5}+1 \times 2^{4}+0 \times 2^{3} \\ & +0 \times 2^{2}+0 \times 2+0 \end{aligned}$$

Therefore, $2640=(101001010000)_{2}$