All of this relies on the isotropy of the uniform distribution of a vector on the unit sphere. Align your coordinate axis with the unit vector $v$, so that $v\cdot a=a_1$. Then use that $\mathbb{E}[a_i^2]$ is independent of the index $i$, so that $m\mathbb{E}[a_1^2]=\sum_{i=1}^m\mathbb{E}[a_i^2]=\mathbb{E}[\sum_{i=1}^m a_i^2]=1$, hence $\mathbb{E}[v\cdot a]=\mathbb{E}[a_1^2]=1/m.$

All of this relies on the isotropy of the uniform distribution of a vector on the unit sphere. Align your coordinate axis with the unit vector $v$, so that $v\cdot a=a_1$. Then use that $\mathbb{E}[a_i^2]$ is independent of the index $i$, so that $m\operatorname{\mathbb{E}} [a_1^2] = \sum_{i=1}^m \mathbb{E}[a_i^2] = \mathbb{E}[\sum_{i=1}^m a_i^2]=1$, hence $\mathbb{E}[v\cdot a]=\mathbb{E}[a_1^2]=1/m.$