### Problem Solution:

#### Part 1:
Given the definition, we need to find the values of $\left(2,8\right)$ and $\left(2, \frac{1}{4}\right)$.

- For $\left(2,8\right)$, we are looking for an $n$ such that $2^{n}=8$. Since $2^{3}=8$, we have $\left(2,8\right)=3$.
- For $\left(2, \frac{1}{4}\right)$, we are looking for an $n$ such that $2^{n}=\frac{1}{4}$. Since $2^{-2}=\frac{1}{4}$, we have $\left(2, \frac{1}{4}\right)=-2$.

Therefore, the answers are: $\boxed{3}$, $\boxed{-2}$.

#### Part 2:
Given $\left(4,12\right)=a$, $\left(4,5\right)=b$, $\left(4,60\right)=c$, we need to prove that $a+b=c$.

- Since $\left(4,12\right)=a$, we have $4^{a}=12$.
- Since $\left(4,5\right)=b$, we have $4^{b}=5$.
- Since $\left(4,60\right)=c$, we have $4^{c}=60$.

Given that $12\times 5=60$, we can write:

$$4^{a}\times 4^{b}=12\times 5=60=4^{c}$$

Using the property that $x^{m}\times x^{n}=x^{m+n}$, we get:

$$4^{a+b}=4^{c}$$

Since the bases are the same, the exponents must be equal, thus:

$$a+b=c$$

Therefore, we have proven that $\boxed{a+b=c}$.