= Solution
Choose a short simple root $\alpha_1$ and a long simple root $\alpha_2$. The <G2 root system> has positive roots
$$
\alpha_1,\ \alpha_2,\ \alpha_1+\alpha_2,\ 2\alpha_1+\alpha_2,\ 3\alpha_1+\alpha_2,\ 3\alpha_1+2\alpha_2.
$$
The two hexagons formed by the short and long roots give the usual twelve-root diagram. The fundamental weights are
$$
\omega_1=2\alpha_1+\alpha_2,\qquad
\omega_2=3\alpha_1+2\alpha_2.
$$
The second is the highest root, so the irreducible module $L(\omega_2)$ is the <Adjoint representation of a Lie algebra>.
The seven weights of $L(\omega_1)$ are zero and the six short roots, each with multiplicity one. Its <crystal basis>[crystal], with arrows denoting the lowering operators $\widetilde f_i$, is the chain
$$
\omega_1
\xrightarrow{1}\alpha_1+\alpha_2
\xrightarrow{2}\alpha_1
\xrightarrow{1}0
\xrightarrow{1}-\alpha_1
\xrightarrow{2}-(\alpha_1+\alpha_2)
\xrightarrow{1}-\omega_1.
$$
Applying the <Weyl dimension formula> to $\lambda=n_1\omega_1+n_2\omega_2$ gives
$$
\dim L(\lambda)=\frac1{120}
(n_1+1)(n_2+1)(n_1+n_2+2)(n_1+2n_2+3)
(n_1+3n_2+4)(2n_1+3n_2+5).
$$
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