(a) The wording here is backwards. One doesn't begin with "for any $(ij)\in G_a$..." Instead, one begins with "for any $i,j\in \{1,\cdots,n\}-\{a\}$" (followed by "$(ij)$ sends $i$ to $j$").
(b) For case II you have a good idea but a few major misunderstandings.
The statement of double transitivity is that given any $i,j\in\{1,\cdots,n\}-\{a\}$, there exists a $\pi\in G_a$ with $\pi(i)=j$. However, there is nothing that says $\pi$ is specifically the transposition $(ij)$, it could be any of the many permutations that send $i$ to $j$. Indeed, if $G$ contained every transposition $(ij)$ then it would be the full symmetric group $S_n$.
If $B$ and $B'$ are blocks of an action of $G$, it's perfectly possible for an element $g\in G$ to send an element of $B$ to an element of $B'$. In particular, in your reasoning it does not follow that $(ij)B$ is not disjoint from $B$. For example, say we have two sets $X$ and $Y$ of the same size. We can consider a group $G$ comprised of permutations of $X\sqcup Y$ with either $f(X)=X$ and $f(Y)=Y$ or else $f(X)=Y$ and $f(Y)=X$. Then $X$ and $Y$ are blocks and there are elements of $G$ that send elements from one block to another.
The exercise is not asking you to argue $G_a$ acts primitively on $\{1,\cdots,n\}-\{a\}$ (indeed, that action may not be primitive). It's asking you to argue $G$ acts on $\{1,\cdots,n\}$ primitively.
So here's how you get your idea to work.
Suppose there is a nontrivial partition. Let $B$ be a block of $\{1,\cdots,n\}$. Then $|B|\ge2$, since if $|B|=1$ it would be a singleton, then applying elements of $G$ gives every other singleton subset (since $G$ acts transitively), telling us this is a trivial partition, a contradiction.
Pick two elements $a,a'\in B$ and another element $b\not\in B$. Then, because $G$ acts doubly transitively, there is a $\pi\in G$ with $\pi(a)=a$ and $\pi(a')=b$. Therefore $\pi(B)$ intersects $B$ nontrivially (since both contain $a$) but is not equal to it (since $\pi(B)$ contains $b$ whereas $B$ doesn't), a contradiction.
Here is some extra commentary to make sense of the notion of primitivity.
First, some elementary set theory. The notions of equivalence relations, set partitions, and onto functions are all more or less equivalent:
- Given an equivalence relation $\sim$ on a set $X$, we get a set partition $X/\sim$ of $X$ into equivalence classes, and an onto function $X\to X/\sim$ which sends an element to its equivalence class.
- Given a set partition $\Gamma$ of a set $X$, we can create an equivalence relation $\sim$ defined by $x\sim y$ if and only if $x,y$ are in the same cell $\gamma\in \Gamma$, and we get an onto function $X\to\Gamma$ where an element is sent to the unique cell it resides in.
- Given an onto function $f:X\to Y$, we get an equivalence relation $\sim$ defined by $x\sim x'$ whenever $f(x)=f(y)$, and $X$ may be partitioned into fibers (preimages of singletons, i.e. the subsets $f^{-1}(y)\subseteq X$ for $y\in Y$).
If you're familiar with the idea of a category, for a given group $G$, there is a category of $G$-sets. If you're not familiar, don't worry about it. Given two $G$-sets $X,Y$, a $G$-set homomorphism is a map $f:X\to Y$ which respects the group action, that is $f(gx)=gf(x)$ for all $x\in X$ and $g\in G$. (One also describes $f$ as equivariant or intertwining.)
A congruence relation $\sim$ on a $G$-set $X$ is an equivalence relation $\sim$ with the added condition that $x\sim y$ implies $gx\sim xy$ for all $x,y\in X$ and $g\in G$. Given any subset $A\subseteq X$, one may apply $g\in G$ to it pointwise, yielding $gA=\{ga:a\in A\}$. A $G$-stable partition of $X$ is a set partition $\Gamma$ with the property that $\gamma\in \Gamma$ implies $g\gamma\in\Gamma$.
Just as in the set-theoretic situation, there is an equivalence between congruence relations on $X$, $G$-stable partitions of $X$, and onto $G$-set homomorphisms out of $X$. Then the primitivity condition on $X$, usually stated in terms of partitions (where the cells are called blocks), can be translated to the statement that $X$ has no nontrivial homomorphic images. Thus, they are the analogue of simple groups in the category of $G$-sets.
For the first one:
You just have to show, that $gx \neq y$ for all $g \in G_y$ and $x \in X \setminus \{y\}$ since all group action axioms are already fullfilled from $G_y \subseteq G$. I hope you can do this (try it! (really!) If you get stuck, post a comment).
Towards the second one:
First consider that you have always at least 2 orbits, since the orbit of $(x,x) \in X \times X$ has always(!) the same two elements in each component (make this yourself clear) and since you have two different elements $x,y \in X$ the element $(x,y)$ is not in this orbit.
For one direction you have to show two orbits are enough, if every $G_y$ is transitive on $X \setminus \{y\}$:
First show, that $G$ is transitive on $X$ (meaning there is only one orbit). Here it is sufficient, that $G$ does not have a fixpoint $y$ (because then there exists $g \in G$ with $gy\neq y$ and because $G_y$ is transitive on $X \setminus \{y\}$ the orbit of $y$ is already everything). Because you have at least 3 elements $x,y,z$ and each element is contained in $X \setminus \{p\}$ for some $p$ you can do this. But this shows, that $\{(x,x) \ | \ x \in X\}$ is already one orbit.
Now you have to show, that $\{(x,y) \ | \ x,y \in X, x \neq y \}$ is an orbit. For fixed $y$ and since $G_y$ acts transitive you can see, that $\{(z,y) \ | \ z \in X, z \neq y\}$ is in the orbit of $(x,y)$ for any $y \neq x \in X$. And samewise for fixed $x \in X \{(x,z) \ | \ z \in X z \neq x\}$is contained in the orbit of $(x,y)$ for any $x \neq y \in X$. Since $x,y$ where arbitrary you can deduce that $\{(x,y) \ | \ x,y \in X, x \neq y \}$ lies in an orbit of $(x,y)$ and since these are now every elements of $X \times X$ you are done.
For the direction backwards consider some $G_y$ which is not transitive and the orbit of $(x,y)$ for some $x \in X$. Now show, that you don't get all elements of this form and hence you need $\geq 3$ different orbits.
Best Answer
If $G$ satisfies 2., and all four of $x_1,x_2, y_1,y_2\in X$ are distinct, then let $g_1$ be an element in $\operatorname{Stab}_G(x_1)$ with $g_1x_2=y_2$ and $g_2\in\operatorname{Stab}_G(y_2)$ with $g_2x_1=y_1$. Then look at $g=g_2g_1$.
If, for instance, $x_1=y_2$, you need to revise the above, but the idea still works.
Edit: Even simpler, and without having to care for special cases: let $g_1\in G$ be any element with $g_1x_1=y_1$, and let $g_2\in \operatorname{Stab}_G(y_1)$ be such that $g_2(g_1x_2)=y_2$. Define $g$ as above.