Given the function $g(x) = x^2 + 4x + 1$, let's solve the following:

1. $g(2)$

Substitute $x = 2$ into the function:

$g(2) = (2)^2 + 4(2) + 1 = 4 + 8 + 1 = 13$

So, $g(2) = 13$.

2. $g(T)$

Substitute $x = T$ into the function:

$g(T) = T^2 + 4T + 1$

So, $g(T) = T^2 + 4T + 1$.

3. $g(x + 2) - x^2 + 4x + 1$

First, we need to find $g(x+2)$. As calculated earlier:

$g(x+2) = x^2 + 8x + 13$

Now, subtract $x^2 + 4x + 1$ from $g(x+2)$:

$g(x+2) - (x^2 + 4x + 1) = (x^2 + 8x + 13) - (x^2 + 4x + 1)$

Simplify the expression by canceling like terms:

$= x^2 + 8x + 13 - x^2 - 4x - 1 = 4x + 12$

So, $g(x+2) - (x^2 + 4x + 1) = 4x + 12$.

Summary:

1. $g(2) = 13$
2. $g(T) = T^2 + 4T + 1$
3. $g(x+2) - x^2 + 4x + 1 = 4x + 12$

Would you like further details or explanations on any of these steps?

Here are five related questions to consider:

1. What is the derivative of $g(x) = x^2 + 4x + 1$ with respect to $x$?
2. How would you solve $g(x) = 0$ for $x$?
3. What does the expression $g(x+2)$ represent geometrically?
4. How does the subtraction $g(x+2) - x^2 + 4x + 1$ relate to the original function?
5. Can $g(x) = x^2 + 4x + 1$ be factored, and if so, how?

Tip: When working with functions, always carefully expand and simplify expressions step by step to avoid mistakes.