Solve in English

Substitute $x=2$ into the function $f(x) = 3x+7$.

The value of $f(2)$ is $\boxed{13}$.

The function $f(x) = 3x+7$ is a linear polynomial function.
Polynomial functions are defined for all real numbers.
The domain of $f(x)$ is $\boxed{\mathbb{R} \text{ or } (-\infty, \infty)}$.

Substitute $x=4$ into the function $g(x) = \frac{1}{x}$.

The value of $g(4)$ is $\boxed{\frac{1}{4}}$.

The composite function $gf(x)$ means $g(f(x))$. We substitute $f(x)$ into $g(x)$.

Since $f(x) = 3x+7$, we have:

Now, substitute $(3x+7)$ into $g(x) = \frac{1}{x}$:

The composite function is $\boxed{gf(x) = \frac{1}{3x+7}}$.

For $gf(x) = \frac{1}{3x+7}$ to be defined, the denominator cannot be zero.

The domain of $gf(x)$ is all real numbers except $x = -\frac{7}{3}$.
The domain of $gf(x)$ is $\boxed{\mathbb{R} \setminus \left\{-\frac{7}{3}\right\} \text{ or } \left(-\infty, -\frac{7}{3}\right) \cup \left(-\frac{7}{3}, \infty\right)}$.

Let $y = 3x+7$.

Subtract 7 from both sides:

Divide by 3:

So, the inverse function is $f^{-1}(x) = \frac{x-7}{3}$.
The inverse function is $\boxed{f^{-1}(x) = \frac{x-7}{3}}$.