= Solution
Choose positive roots
$$
\Phi^+=\{\varepsilon_1-\varepsilon_2,\varepsilon_2,
\varepsilon_1,\varepsilon_1+\varepsilon_2\}.
$$
The corresponding simple roots are $\varepsilon_1-\varepsilon_2$ and $\varepsilon_2$. The vector $v_{1,+}\wedge v_{2,+}$ has weight $\lambda=\varepsilon_1+\varepsilon_2$. Adding any positive root to this weight produces a weight absent from part (a), so every positive root vector annihilates it. It is consequently a <highest-weight vector> with highest weight $\lambda$.
A representation of a <compact Lie group> is completely reducible: Haar averaging makes an inner product invariant, and orthogonal complements of invariant subspaces remain invariant. Therefore the <exterior square> contains an irreducible summand $L_\lambda$ of this highest weight. This can also be seen by projecting the displayed highest-weight vector into irreducible summands; a nonzero projection remains annihilated by all positive root vectors and has the same weight. Its multiplicity is one because the weight space at $\lambda$ is one-dimensional.
Use the <Weyl dimension formula>, expressly allowed in this part:
$$
\dim L_\lambda=\prod_{\alpha\in\Phi^+}
\frac{\langle\lambda+\rho,\alpha\rangle}{\langle\rho,\alpha\rangle}.
$$
Here $L_\lambda$ is the irreducible highest-weight representation, $\Phi^+$ is the chosen positive root system, $\rho$ is half its sum, and the pairing is the invariant Euclidean inner product on the real weight plane, with $\varepsilon_1,\varepsilon_2$ orthonormal. Using coroots instead gives the same ratio because the length factor cancels for each root. In this convention,
$$
\rho=\tfrac32\varepsilon_1+\tfrac12\varepsilon_2,
\qquad\lambda+\rho=\tfrac52\varepsilon_1+\tfrac32\varepsilon_2.
$$
The four factors, in the listed positive-root order, are respectively
$$
\frac{1}{1},\qquad\frac{3/2}{1/2},\qquad
\frac{5/2}{3/2},\qquad\frac{4}{2}.
$$
Their product is $10$. Since $\dim\Lambda^2\mathbb C^5=\binom52=10$, this irreducible summand occupies the entire exterior square. Hence
$$
\boxed{\Lambda^2\mathbb C^5\text{ is irreducible as an }SO(5)\text{-representation}.}
$$
The proof uses the dimension formula only where it is allowed, not in the symmetric-power argument of Question 2. Its <B2 adjoint representation weight diagram> is exactly the eight roots and the double zero weight obtained above.
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