Let me analyze the math content in the image step by step. I'll solve each question and provide detailed explanations. Here's the breakdown of what I can see:

Problem 1: Simplify or calculate the following expressions

1. (a) $-32.56 - 32.25 - 32.19$
   Combine terms directly: $-32.56 - 32.25 - 32.19 = -(32.56 + 32.25 + 32.19) = -97.00$

2. (b) $\frac{9^3 \cdot 25^3}{18^2 \cdot 125^2}$
   Simplify the powers and factors:

   • $9 = 3^2$ and $25 = 5^2$, so $9^3 = 3^6$ and $25^3 = 5^6$.
   • $18 = 2 \cdot 3^2$, so $18^2 = 2^2 \cdot 3^4$, and $125 = 5^3$, so $125^2 = 5^6$.
     The expression becomes: $\frac{3^6 \cdot 5^6}{2^2 \cdot 3^4 \cdot 5^6} = \frac{3^{6-4} \cdot 5^{6-6}}{2^2} = \frac{3^2}{2^2} = \frac{9}{4}.$

3. (c) $\frac{(3 \cdot 4 \cdot 2^{16})^2}{11 \cdot 12^{13} \cdot 4^{11} \cdot 16^9}$
   Rewrite terms in powers:

   • $4 = 2^2$, $16 = 2^4$, so $4^{11} = 2^{22}$ and $16^9 = 2^{36}$.
   • $(3 \cdot 4 \cdot 2^{16})^2 = 3^2 \cdot 4^2 \cdot 2^{32} = 3^2 \cdot 2^4 \cdot 2^{32} = 3^2 \cdot 2^{36}$.
     The denominator becomes: $11 \cdot 12^{13} \cdot 4^{11} \cdot 16^9 = 11 \cdot (2^2 \cdot 3)^{13} \cdot 2^{22} \cdot 2^{36} = 11 \cdot 2^{26} \cdot 3^{13} \cdot 2^{36}.$ Combining terms: $\frac{3^2 \cdot 2^{36}}{11 \cdot 2^{58} \cdot 3^{13}} = \frac{3^{2-13} \cdot 2^{36-58}}{11} = \frac{3^{-11} \cdot 2^{-22}}{11}.$

   Simplify further if needed.

Problem 2: Compare $A$ and $B$

• $A = 1 + 2012 + 2012^2 + \ldots + 2012^{72}$, which is a geometric series with first term $a = 1$, common ratio $r = 2012$, and $n = 73$ terms.
  The sum of a geometric series is: $A = \frac{1 \cdot (2012^{73} - 1)}{2012 - 1}.$

• $B = 2012^{73} - 1$.

Clearly, $A < B$ because $A$ is a fraction of $B$.

Problem 3: Compare $E$ and $Q$

• $E = \frac{2018^{99} - 1}{2018^{100} - 1}$.
  Divide numerator and denominator by $2018^{99}$: $E = \frac{1 - \frac{1}{2018^{99}}}{1 - \frac{1}{2018^{100}}}.$

• $Q = \frac{2018^{98} - 1}{2018^{99} - 1}$.
  Similarly: $Q = \frac{1 - \frac{1}{2018^{98}}}{1 - \frac{1}{2018^{99}}}.$

Since $2018^{99} > 2018^{98}$, it follows that $E > Q$.

Problem 4: Find prime numbers $p$ such that $p + 8$ and $p + 10$ are also prime.

Let $p$ be a prime number. For $p + 8$ and $p + 10$ to also be prime:

• $p \neq 2$ (since $p+8$ would be even and not prime).
• Test small odd primes $p$:
  • $p = 3$: $3+8 = 11$ (prime), $3+10 = 13$ (prime).
  • $p = 5$: $5+8 = 13$ (prime), $5+10 = 15$ (not prime).

Thus, $p = 3$ works.

Would you like more detailed explanations or solutions for any of these problems?

Here are 5 related questions to explore further:

1. What is the general formula for simplifying geometric series?
2. How can we systematically find prime numbers that satisfy multiple conditions?
3. Can $E$ and $Q$ in Problem 3 be expressed as limits as $n \to \infty$?
4. What patterns exist for primes $p$ where $p + k$ (for fixed $k$) is also prime?
5. How do properties of exponents simplify expressions like those in Problem 1(b)?

Tip: For simplifying powers and factorial-like terms, factorize numbers into primes early—it helps to cancel terms efficiently!