This language was already considered by the same OP in another
question Build a regular expression to define a regular language. But then, posters are told not to ask two questions in the same post.
I already answered that a simple method is to consider the regular
languages defined by conditions 1 and 2, $1a(1+a+b)^*$, and by
conditions 1 and 3, $(1+a+b)^*aa(1+a+b)^*$.
You take the FA for them, both having 3 states (4 if you count a dead state), and construct the FA
for the intersection. Which is quite easy, and gives a five-states
automaton (6 states if you count a dead state).
From that FA you can get a regular expression for the language, which is
But you can also get a regular grammar:
$S \rightarrow 1T$
$T \rightarrow aX \mid aY$
$X \rightarrow 1X \mid aX \mid bX \mid aY$
$Y \rightarrow a \mid aZ$
$Z \rightarrow 1 \mid a \mid b \mid 1Z \mid aZ \mid bZ$
which derives directly from the automaton.
In more details: Why do it as I did above?
Proving that a regular expression defines the language specified by a
set of such conditions can be long and tedious. The same is true for a
grammar or an automaton. This is why the proper way of answering such
a question is to find a systematic way of building the desired result
from elementary components for which things are obvious to prove.
Here, as I had already answered, elementary components are the regular
languages defined respectively
For each of these 2 languages, it is easy to find the regular
expression (above) or to give a regular grammar or a finite-state
automaton (FA). Each has a 3 states FA that recognizes it (4 states including a dead state).
To get a FA for the language meeting all three conditions, you only
have to get the intersection of these two languages. There is a
standard cross-product construction for building a FA
recognizing the intersection of two regular languages. This
construction is easy to apply by hand in our example. It produce a 5
states FA for the language defined by all three conditions (or a 6 states FA, including a dead state).
From this last FA, it is easy to build either the regular expression
or the regular grammar given above.
Since the answer was built in a systematic way using well known
techniques (that have been proved correct long ago), there is no need
to prove that the resulting regular expression, grammar or FA are
correct. Barring of course mistakes in the use of the constructions
... but then, you can also make mistakes in proofs.
I added the precision about dead states so that no one gets confused. But, except for some constructions, it is often simpler just to omit dead states and transitions leading to them.