To prove the statements given \( \frac{a}{b} = \frac{c}{d} \), we can utilize the property of proportions.

### Given:
\[
\frac{a}{b} = \frac{c}{d}
\]
This implies:
\[
a \cdot d = b \cdot c
\]

### a) To prove:
\[
(a + 2c)(b + d) = (a + c)(b + 2d)
\]
**Proof:**

Expanding both sides:

1. Left-hand side:
\[
(a + 2c)(b + d) = ab + ad + 2bc + 2cd
\]

2. Right-hand side:
\[
(a + c)(b + 2d) = ab + 2ad + bc + 2cd
\]

Now we compare:
- From LHS: \( ab + ad + 2bc + 2cd \)
- From RHS: \( ab + 2ad + bc + 2cd \)

Setting them equal requires simplifying:
\[
ad + 2bc = 2ad + bc \implies 2bc - bc = 2ad - ad \implies bc = ad
\]

Since \( a \cdot d = b \cdot c \), this holds true.

### b) To prove:
\[
\frac{a^{2000} + b^{2000}}{c^{2000} + d^{2000}} = \left( \frac{a+b}{c+d} \right)^{2000}
\]

**Proof:**

By the property of homogenous polynomials, we know:

\[
\frac{x^n + y^n}{z^n + w^n} = \left( \frac{x + y}{z + w} \right)^n \quad (\text{for any positive integer } n)
\]

In our case:
- \( x = a \)
- \( y = b \)
- \( z = c \)
- \( w = d \)

Since both \( a \) to \( b \) and \( c \) to \( d \) maintain the same ratio, thus:
\[
\frac{a + b}{c + d} = k \quad (\text{some constant})
\]

Raising it to the 2000th power preserves the equality. Therefore, the property holds true:
\[
\frac{a^{2000} + b^{2000}}{c^{2000} + d^{2000}} = \left( \frac{a + b}{c + d} \right)^{2000}
\]

### Conclusion:
Both statements are confirmed, demonstrating the relationships under the given condition \( \frac{a}{b} = \frac{c}{d} \).