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The image contains a set of mathematical expressions that need to be simplified. Let's solve each part step by step:

**62. Apskaičiuokite (Calculate):**

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**a) \( 2^2 : 4^2 \)**  
We simplify:
- \( 2^2 = 4 \)
- \( 4^2 = 16 \)

Thus:
\[ \frac{2^2}{4^2} = \frac{4}{16} = \frac{1}{4} \]

Answer: \( \frac{1}{4} \)

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**b) \( 2^1 : 16^2 \)**  
First, express \( 16 \) as a power of 2:
- \( 16 = 2^4 \), so \( 16^2 = (2^4)^2 = 2^8 \)

Now simplify:
\[ \frac{2^1}{2^8} = 2^{1-8} = 2^{-7} = \frac{1}{2^7} = \frac{1}{128} \]

Answer: \( \frac{1}{128} \)

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**c) \( 25^1 : 5^2 \)**  
First, express \( 25 \) as \( 5^2 \):
- \( 25^1 = 5^2 \)

Now simplify:
\[ \frac{5^2}{5^2} = 1 \]

Answer: \( 1 \)

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**d) \( 81^2 : 3^5 \)**  
First, express \( 81 \) as \( 3^4 \):
- \( 81^2 = (3^4)^2 = 3^8 \)

Now simplify:
\[ \frac{3^8}{3^5} = 3^{8-5} = 3^3 = 27 \]

Answer: \( 27 \)

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**e) \( 2^5 \cdot 8^3 \cdot 32^2 \)**  
Express \( 8 \) and \( 32 \) as powers of 2:
- \( 8 = 2^3 \), so \( 8^3 = (2^3)^3 = 2^9 \)
- \( 32 = 2^5 \), so \( 32^2 = (2^5)^2 = 2^{10} \)

Now multiply:
\[ 2^5 \cdot 2^9 \cdot 2^{10} = 2^{5+9+10} = 2^{24} \]

Answer: \( 2^{24} \)

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**f) \( 81^1 \cdot 27^2 \cdot 3^5 \)**  
Express \( 81 \) and \( 27 \) as powers of 3:
- \( 81 = 3^4 \), so \( 81^1 = 3^4 \)
- \( 27 = 3^3 \), so \( 27^2 = (3^3)^2 = 3^6 \)

Now multiply:
\[ 3^4 \cdot 3^6 \cdot 3^5 = 3^{4+6+5} = 3^{15} \]

Answer: \( 3^{15} \)

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**g) \( \frac{81^2}{12^2 \cdot 4^3} \)**  
First, express everything as powers:
- \( 81 = 3^4 \), so \( 81^2 = (3^4)^2 = 3^8 \)
- \( 12 = 2^2 \cdot 3 \), so \( 12^2 = (2^2 \cdot 3)^2 = 2^4 \cdot 3^2 \)
- \( 4 = 2^2 \), so \( 4^3 = (2^2)^3 = 2^6 \)

Now simplify:
\[ \frac{3^8}{(2^4 \cdot 3^2) \cdot 2^6} = \frac{3^8}{2^{4+6} \cdot 3^2} = \frac{3^8}{2^{10} \cdot 3^2} = \frac{3^{8-2}}{2^{10}} = \frac{3^6}{2^{10}} \]

Answer: \( \frac{3^6}{2^{10}} \)

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**h) \( \frac{2^4 \cdot 16^2}{4^3 \cdot 2^1} \)**  
First, express \( 16 \) and \( 4 \) as powers of 2:
- \( 16 = 2^4 \), so \( 16^2 = (2^4)^2 = 2^8 \)
- \( 4 = 2^2 \), so \( 4^3 = (2^2)^3 = 2^6 \)

Now simplify:
\[ \frac{2^4 \cdot 2^8}{2^6 \cdot 2^1} = \frac{2^{4+8}}{2^{6+1}} = \frac{2^{12}}{2^7} = 2^{12-7} = 2^5 \]

Answer: \( 2^5 \)

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**i) \( \frac{3^5 \cdot 27}{81 \cdot 3^5} \)**  
Express \( 27 \) and \( 81 \) as powers of 3:
- \( 27 = 3^3 \)
- \( 81 = 3^4 \)

Now simplify:
\[ \frac{3^5 \cdot 3^3}{3^4 \cdot 3^5} = \frac{3^{5+3}}{3^{4+5}} = \frac{3^8}{3^9} = 3^{8-9} = 3^{-1} = \frac{1}{3} \]

Answer: \( \frac{1}{3} \)

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Would you like more details on any of these solutions, or do you have any questions?

### Related Questions:
1. What are the general laws of exponents used in simplifying these expressions?
2. How can fractional exponents be simplified in similar problems?
3. Can negative exponents appear in real-world applications, and if so, how?
4. How do expressions with different bases get simplified, such as \( 2^x \) and \( 3^y \)?
5. What are the common mistakes to avoid when simplifying expressions involving powers and roots?

### Tip:
Always break down large expressions into smaller parts, using the properties of exponents step-by-step to avoid confusion.

62. Apskaičiuokite: a) 2^2 : 4^2, b) 2^1 : 16^2, c) 25^1 : 5^2, d) 81^2 : 3^5, e) 2^5 ⋅ 8^3 ⋅ 32^2, f) 81^1 ⋅ 27^2 ⋅ 3^5, g) 81^2 ÷ (12^2 ⋅ 4^3), h) (2^4 ⋅ 16^2) ÷ (4^3 ⋅ 2^1), i) (3^5 ⋅ 27) ÷ (81 ⋅ 3^5)

Learn how to simplify complex exponent expressions from problem 62. The problem involves simplifying expressions with powers, using exponent laws, and applying division and multiplication rules. Suitable for students in Grades 9-11.

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