Artin, Algebra, Chapter 3, Ex. 4.4
I can prove (b), viz., that
The order of $GL_2(\mathbb{F}_p)=p(p+1)(p-1)^2$
The order of $SL_2(\mathbb{F}_p)=p(p+1)(p-1)$
However, I have no idea how to prove (a), viz.,
Why the number of bases of $V=\mathbb{F}_p^2$ is equal to the order of $GL_2(\mathbb{F}_p)$?
Fix an ordered basis $B_0$. If $T$ is an invertible linear transformation then the image of $B_0$ under $T$ is another ordered basis. And given another ordered basis $B$ it is the image of $B_0$ under exactly one such $T$.
Consider the map $\phi$ from the ordered bases of $V$ to $GL_2(\mathbb{F}_p)$ given by $\phi(v_1,v_2)=[v_1,v_2]$. The map is surjective, since the columns of an invertible matrix form an ordered basis, and is also trivially injective. Hence $\phi$ is a bijection and the result follows.
For any field $K$, a matrix $A\in K^{n\times n}$ is invertible if and only its column form a basis of $K^n$. This means you have a bijection between the set of basis of $K^n$ and the set of invertible matrix $GL_n(K)$.
If $K$ is finite, these sets are also finite and thus have the same number of elements.
