Complete the missing parts of algebraic multiplication and division problems.

To solve the given problems involving algebraic multiplication and division, we will apply the rules of exponents and basic algebraic operations. Let's go through each part step by step.

1. Multiplication Problems

#### (a) \( a^3 \times ab \)
- When multiplying terms with the same base, add the exponents.
- Here, \( a^3 \times ab = a^3 \times a^1 \times b^1 \).
- Combine the exponents for \( a \): \( 3 + 1 = 4 \).
- The term \( b \) remains as \( b^1 \).

So, \( a^3 \times ab = a^4b \).

#### (b) \( 2a \times 3ab \)
- Multiply the coefficients: \( 2 \times 3 = 6 \).
- For the variables, combine the exponents:
- For \( a \): \( a^1 \times a^1 = a^{1+1} = a^2 \).
- The term \( b \) remains as \( b^1 \).

So, \( 2a \times 3ab = 6a^2b \).

#### (c) \( x^2y \times \_\_ = 2x^2y^3 \)
- We need to find the missing term such that when multiplied by \( x^2y \), the result is \( 2x^2y^3 \).
- Compare the coefficients and exponents:
- Coefficient: The result has a coefficient of 2, so the missing term must have a coefficient of 2.
- For \( x \): The exponent of \( x \) in the result is 2, which matches \( x^2 \) in \( x^2y \). So, the missing term does not contribute to \( x \).
- For \( y \): The exponent of \( y \) in the result is 3, but \( x^2y \) contributes \( y^1 \). Therefore, the missing term must contribute \( y^{3-1} = y^2 \).

So, the missing term is \( 2y^2 \).

#### (d) \( 3d^4 \times \_\_ = 6a^2d^5 \)
- We need to find the missing term such that when multiplied by \( 3d^4 \), the result is \( 6a^2d^5 \).
- Compare the coefficients and exponents:
- Coefficient: The result has a coefficient of 6, so the missing term must have a coefficient of \( \frac{6}{3} = 2 \).
- For \( d \): The exponent of \( d \) in the result is 5, and \( 3d^4 \) contributes \( d^4 \). Therefore, the missing term must contribute \( d^{5-4} = d^1 \).
- For \( a \): The result has \( a^2 \), but \( 3d^4 \) does not contribute any \( a \). Therefore, the missing term must contribute \( a^2 \).

So, the missing term is \( 2a^2d \).

2. Division Problems

#### (a) \( 6ab \div 2 \)
- Divide the coefficients: \( \frac{6}{2} = 3 \).
- The variables \( a \) and \( b \) remain unchanged because they are not divided by any variable.

So, \( 6ab \div 2 = 3ab \).

#### (b) \( a^2b \div a^2 \)
- Divide the coefficients: The coefficients are both 1 (implied), so the result is 1.
- For the variables:
- For \( a \): \( \frac{a^2}{a^2} = a^{2-2} = a^0 = 1 \).
- The term \( b \) remains as \( b^1 \).

So, \( a^2b \div a^2 = b \).

#### (c) \( ab^3 \div ab \)
- Divide the coefficients: The coefficients are both 1 (implied), so the result is 1.
- For the variables:
- For \( a \): \( \frac{a^1}{a^1} = a^{1-1} = a^0 = 1 \).
- For \( b \): \( \frac{b^3}{b^1} = b^{3-1} = b^2 \).

So, \( ab^3 \div ab = b^2 \).

#### (d) \( 4a^2b \div \_\_ = 2a^2 \)
- We need to find the missing divisor such that when \( 4a^2b \) is divided by it, the result is \( 2a^2 \).
- Compare the coefficients and exponents:
- Coefficient: The result has a coefficient of 2, so the missing divisor must have a coefficient of \( \frac{4}{2} = 2 \).
- For \( a \): The exponent of \( a \) in the result is 2, which matches \( a^2 \) in \( 4a^2b \). So, the missing divisor does not contribute to \( a \).
- For \( b \): The result has no \( b \), but \( 4a^2b \) contributes \( b^1 \). Therefore, the missing divisor must be \( b^1 \).

So, the missing divisor is \( 2b \).

Final Answers

\[
\boxed{
\begin{array}{|c|c|c|}
\hline
a^3 \times ab & = a^4b & 6ab \div 2 = 3ab \\
2a \times 3ab & = 6a^2b & a^2b \div a^2 = b \\
x^2y \times 2y^2 & = 2x^2y^3 & ab^3 \div ab = b^2 \\
3d^4 \times 2a^2d & = 6a^2d^5 & 4a^2b \div 2b = 2a^2 \\
\hline
\end{array}
}
\]