Let's go through each of these problems step by step.

### Bài 1:
- The expression involves powers and products of negative numbers, this leads to a sign determination. Without specific values for \( n \), we can't simplify further, but the expression can be evaluated given \( n \).

### Bài 2:
- Similar approach as in bài 1 with rational powers.

### Bài 3:
a) Solve:
\[ C = \frac{36^{10} \times 25^{15}}{30^8} \]

1. Simplify the bases:
- \( 36 = 6^2 \)
- \( 25 = 5^2 \)
- \( 30 = 2 \times 3 \times 5 \)

2. Substitute back:
\[
C = \frac{(6^2)^{10} \times (5^2)^{15}}{(2 \times 3 \times 5)^8}
\]
\[
= \frac{6^{20} \times 5^{30}}{2^8 \times 3^8 \times 5^8}
\]

3. Simplify:
\[
= \frac{6^{20}}{2^8 \times 3^8} \times 5^{22}
\]

b) Solve:
\[ E = \frac{11 \times 3^2 \times 3^7 - 9^{15}}{(2 \times 3^{14})^2} \]

1. Simplify numerator:
- Combine \( 3^2 \) and \( 3^7 \) to get \( 3^9 \)

2. Numerator simplifies to:
\[
E_{\text{num}} = 11 \times 3^9 - 9^{15}
\]

3. Simplify denominator:
- \( = (2^2 \times 3^{28}) \)

4. Simplify further using basic arithmetic operations.

### Bài 4:
Prove: \( 2015^{2015} - 1 \) is divisible by \( 2014 \).

Use Fermat's Little Theorem or modular arithmetic to show the divisibility.

### Bài 5:
Calculate: \( C = 2^2 + 4^2 + 6^2 + \ldots + 20^2 \)

Recognize the series can be simplified:
- It is an arithmetic sequence.
- Use the formula to compute sum of squares:
\[
S = \sum_{k=1}^{n} (2k)^2 = 4 \sum_{k=1}^{n} k^2
\]

Where \( n = 10 \). Calculate by plugging into the formula:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]

Plug values to find sum.