2. \frac{(2^{17}) \cdot (3^{4})}{(24^{4})}= \frac{(2^{17}) \cdot (3^{4})}{(3 \cdot \Box)^{4}}= \frac{(2^{17}) \cdot (3^{4})}{(3^{4} \cdot \Box)}= \frac{(2^{17})}{(\Box)^{4}}= \frac{(2^{17})}{(2^{12})}= 2^{\Box} =\Box

2. Решим пример:

$$\frac{(2^{17}) \cdot (3^{4})}{(24^{4})}=\frac{(2^{17}) \cdot (3^{4})}{(3 \cdot 8)^{4}}=\frac{(2^{17}) \cdot (3^{4})}{(3^{4} \cdot 8^4)}=\frac{(2^{17})}{(8)^{4}}=\frac{(2^{17})}{(2^3)^{4}}=\frac{(2^{17})}{(2^{12})}= 2^{(17-12)}=2^5=32$$.

Заполним пропуски:

$$\frac{(2^{17}) \cdot (3^{4})}{(24^{4})}=\frac{(2^{17}) \cdot (3^{4})}{(3 \cdot \boxed{8})^{4}}=\frac{(2^{17}) \cdot (3^{4})}{(3^{4} \cdot \boxed{8^4})}=\frac{(2^{17})}{(\boxed{8})^{4}}=\frac{(2^{17})}{(2^{12})}= 2^{\boxed{5}} =\boxed{32}$$.

Ответ: 8; 8⁴; 8; 5; 32.