= Solution
The <permutation representation> on $\mathbb C^X$ sends the basis vector $e_x$ to $e_{gx}$. Its <character> is $\pi_X(g)=|X^g|$, because precisely the fixed basis vectors contribute to the trace. <Burnside lemma> says that the number of orbits is $|G|^{-1}\sum_g|X^g|$. To prove it, double-count $(g,x)$ with $gx=x$: the count is $\sum_g|X^g|=\sum_x|G_x|$. Each orbit contributes $|Gx||G_x|=|G|$ by the <orbit-stabilizer theorem>.
A pair is fixed under the diagonal action exactly when both entries are fixed. Since permutation characters are real,
$$
\langle\pi_{X_1},\pi_{X_2}\rangle=\frac1{|G|}\sum_g |X_1^g||X_2^g|
=\#(G\backslash(X_1\times X_2)),
$$
by <Burnside lemma> on the product.
The <general linear group> $\operatorname{GL}_2(q)$ acts on the $q+1$ lines of $\mathbb F_q^2$. It is doubly transitive: choosing nonzero representatives of any two distinct lines gives a basis, and a linear map can send that basis to representatives of any other ordered pair. Thus there are two orbits on ordered pairs, the diagonal and its complement. The <character inner product> of this permutation character with itself is two, while the multiplicity of the trivial character is one by transitivity. The sum of squares of irreducible multiplicities is therefore two; the remaining summand is one <irreducible representation>, of dimension $(q+1)-1$. \b[Its dimension is $\boxed q$.]
For $S_n$ acting on two-element subsets, an ordered pair of subsets is classified by intersection size $0,1,2$. All three occur for $n\geq4$, and a permutation carries any pair of a given type to any other. Hence $\langle\pi_Z,\pi_Z\rangle=3$. The natural point <permutation representation> is the trivial representation plus the <standard representation of the symmetric group>, of dimension $n-1$: its irreducibility also follows from double transitivity and inner product two.
Embed this point representation into $\mathbb C^Z$ by $e_i\mapsto\sum_{j\ne i}e_{\{i,j\}}$. The map is equivariant, and its kernel consists of vectors with $c_i+c_j=0$ for every distinct $i,j$. Three distinct indices force every $c_i=0$, so it is injective for $n\geq3$. Thus the trivial and standard representations each occur in $\mathbb C^Z$. The norm three leaves exactly one further irreducible, with multiplicity one. It is the orthogonal complement of that embedded subspace, equivalently the vectors $a_{ij}=a_{ji}$ satisfying $\sum_{j\ne i}a_{ij}=0$ for every $i$. Its dimension is
$$
\binom n2-n=\frac{n(n-3)}2.
$$
\b[The constituent dimensions are $\boxed{1,\ n-1,\ n(n-3)/2}$], each once. For $n=3$, complementation identifies two-element subsets with points, so there are only the trivial and 2-dimensional standard constituents.
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