1. **Problem statement:** (a) Determine between which two consecutive integers the given irrational numbers lie. 2. **Formula and rules:** To find between which two consecutive integers a number $x$ lies, find integers $n$ and $n+1$ such that $n < x < n+1$. 3. **Step-by-step solution for (a):** (1) $\sqrt{8}$: Since $2^2=4$ and $3^2=9$, $\sqrt{8}$ lies between 2 and 3. (2) $\sqrt{18}$: $4^2=16$ and $5^2=25$, so $\sqrt{18}$ lies between 4 and 5. (3) $\sqrt{21}$: $4^2=16$ and $5^2=25$, so $\sqrt{21}$ lies between 4 and 5. (4) $\sqrt{50}$: $7^2=49$ and $8^2=64$, so $\sqrt{50}$ lies between 7 and 8. (5) $-\sqrt[3]{49}$: Cube root of 49 is between $3^3=27$ and $4^3=64$, so $\sqrt[3]{49}$ is between 3 and 4, thus $-\sqrt[3]{49}$ lies between -4 and -3. (6) $-\sqrt{42}$: $6^2=36$ and $7^2=49$, so $\sqrt{42}$ lies between 6 and 7, thus $-\sqrt{42}$ lies between -7 and -6. (7) $-\sqrt{150}$: $12^2=144$ and $13^2=169$, so $\sqrt{150}$ lies between 12 and 13, thus $-\sqrt{150}$ lies between -13 and -12. (8) $\sqrt{74}$: $8^2=64$ and $9^2=81$, so $\sqrt{74}$ lies between 8 and 9. (9) $-\frac{\sqrt[3]{6}}{\pi}$: Cube root of 6 is between $1^3=1$ and $2^3=8$, so $\sqrt[3]{6}$ is between 1 and 2. Since $\pi \approx 3.14$, $\frac{\sqrt[3]{6}}{\pi}$ is between $\frac{1}{3.14} \approx 0.318$ and $\frac{2}{3.14} \approx 0.637$. Thus, $-\frac{\sqrt[3]{6}}{\pi}$ lies between -0.637 and -0.318. (10) $\pi$: $3 < \pi < 4$. (11) $5\pi$: Since $\pi \approx 3.14$, $5\pi \approx 15.7$, so lies between 15 and 16. (12) $-\frac{3}{4}\sqrt{}$ (assuming $-\frac{3}{4}\sqrt{1}$ or incomplete, cannot determine exact value, skipping). 4. **Final answers for (a):** (1) between 2 and 3 (2) between 4 and 5 (3) between 4 and 5 (4) between 7 and 8 (5) between -4 and -3 (6) between -7 and -6 (7) between -13 and -12 (8) between 8 and 9 (9) between -0.637 and -0.318 (10) between 3 and 4 (11) between 15 and 16 5. **For (b) and (c), recurring decimals as fractions and rational numbers:** (1) $0.5 = \frac{5}{9}$ (if recurring 0.555...) or $\frac{1}{2}$ if terminating. (2) $0.27$ recurring means $0.272727... = \frac{27}{99} = \frac{3}{11}$. (3) $0.32$ recurring means $0.323232... = \frac{32}{99}$. (4) $13.55$ recurring means $13.5555... = 13 + \frac{5}{9} = \frac{122}{9}$. (1) $0.215$ recurring means $0.215215215... = \frac{215}{999}$. (2) $4.68\overline{8}$ means $4.688888... = 4 + \frac{8}{90} = \frac{421}{90}$. (3) $-3.124\overline{4}$ means $-3.124444... = -\left(3 + \frac{1244}{9990}\right) = -\frac{31206}{9990}$. (4) $-6.49\overline{9}$ means $-6.499999... = -6.5 = -\frac{13}{2}$. These show the decimals are rational numbers.