What is the intuition and motivation behind the definition of the thermodynamic activity?
So I'm taking Pchem right now so I'm no experts, but here is my two cents.

I'll be honest I essentially followed my books lead on the determination of $a_{i}$, and $\mu_{i}$. When looking at problems like this I try to go backwards. I see that $a_{i}$ is derived from $\mu_{i}$, and I try to find where $\mu_{i}$ comes from then see where that comes from until I get a definition or a well known relation like Gibbs free energy. I'm not savy enough to be like "Oh yea $\mu_{i}$, that's basically $\int dG(T, P_{2})$ I'll just show all that". But, I can slowly go backward. So, I followed the following path.
$$a_{i} \Rightarrow \mu_{i} \Rightarrow (\mu(T,P_{2})-\mu(T,P_{1}) ) \Rightarrow d \mu \Rightarrow d G$$
Then when you have this general path you reverse it to get the derivation.

Also remember that you have to keep track of all the assumptions you make. Its important to remember that. My answer is only applicable for a perfect gas at constant temperature, and a mixture of gas at low pressure. 



From deferential of Gibbs free energy the chemical potential of a pure gas is, 
$$dG = -SdT +VdP = d\mu$$
at constant temperature,
$$d \mu = V_{m}dP = \frac {RT}{P}dP$$
To get $\mu$ we integrate with constant temperature essentially  expanding the gas on an isotherm for $P_{1}$ to $P_{2}$.
$$\int _{1}^{2} d \mu = RT \int _{P_{1}}^{P_{2}} \frac{1}{P}dP$$
$$\mu(T,P_{2})-\mu(T,P_{1})=RT \ln \Big{(}\frac {P_{2}}{P_{1}}\Big{)}$$
Let $\mu(T,P_{1})=\mu^{\circ} $,  $P_{1}=P^{\circ}=1bar$, $P_{2}=P$, and $\mu(T,P_{2})=\mu_{i} $,

$$\mu = \mu_{i}+ RT \ln \Big{(}\frac {P}{P^{\circ}}\Big{)}$$

Side note your equation 3 comes from this equation. $\mu_{i}=G_{i}$.

For a mix of gas at low pressure, the activity is equal to the ratio of the partial pressure of the gas over the standard pressure,
$$a_{i}=\Big{(}\frac{P}{P^{\circ}}\Big{)}$$

$$\mu_{i} = \mu_{i}^{\circ} +RT \ln a_{i}$$

I think that's how you prove that relationship. I hope that helped.
$$\mu_{i} = \mu_{i}^{\circ} +RT \ln a_{i} \Rightarrow a_{i}= e^{\frac{\mu_{i}-\mu^{\circ}}{RT}} $$
This relation is trivial and just some algebra. When you are trying prove something its important to start as far back as possible.



Citations:

Chemical Activity

Levine, Ira N. Physical Chemistry. 6th ed. pg 175-176 . New York: McGraw-Hill, 1978. Print.

Notes from my professor (sorry).