Solution

Given language is:

$L={a^n b^{n-3}\mid n>2}$

When $n=3$:

$a^3b^0=aaa$

When $n=4$:

$a^4b^1=aaaab$

When $n=5$:

$a^5b^2=aaaaabb$

So, every string starts with at least three $a$ symbols.

After that, for every extra $a$, one $b$ is added at the end.

Grammar:

$S\rightarrow aaaA$

$A\rightarrow aAb\mid \lambda$

If $A\rightarrow \lambda$, then string is:

$aaa$

If $A\rightarrow aAb$, then it generates:

$aaaab,\ aaaaabb,\ldots$

So, this grammar generates $a^n b^{n-3}$ for $n>2$.