A person is traveling between two places, and has 3 options for transportation. The jth option would take an average of µj hours, with a standard deviation of $\sigma_j$ hours. The person randomly chooses between the 3 options, with equal probabilities. Let T be how long it takes for him to get from place 1 to place 2.
(a) Find E(T). Is it simply (µ1 +µ2 +µ3)/3, the average of the expectations?
Expectation is additive, there is equal probabilities between options, so I agree it is the "average" of the averages.
(b) Find Var(T). Is it simply ($\sigma_1^2+\sigma_2^2+\sigma_j^2$)/3, the average of the variances?
Variance is additive when you are adding iid E(T)'s (confirmed: https://apcentral.collegeboard.org/courses/ap-statistics/classroom-resources/why-variances-add-and-why-it-matters), but I am unsure if you just "average" the variances. Guidance?
Call $C$ the option chosen, thus $C$ is uniform on $\{1,2,3\}$. The random variable $C$ allows to give a rigorous, pathwise, description of $T$ as
where $X_j$ is the transportation time for option $j$.
Now, add to the pot the crucial assumption that:
That is, one assumes that the passenger chooses their option without any knowledge about the transportation times they shall have to endure (otherwise, one could imagine that $T$ is the minimum of $\{X_1,X_2,X_3\}$...).
With this representation of $T$ and this independence hypothesis, we can compute every characteristic of $T$. For example, by independence, $$E(T)=\sum_jP(C=j)E(X_j)=\frac13\sum_j\mu_j$$ as you predicted (but for a slightly different reason). Likewise, $$T^2=\sum_j\mathbf 1_{C=j}X_j^2$$ hence, again by independence, $$E(T^2)=\sum_jP(C=j)E(X_j^2)=\frac13\sum_j(\sigma_j^2+\mu_j^2)$$ which implies that
More generally, for any given number of choices, each with probability $p_j$, one would get $$E(T)=\sum_jp_j\mu_j$$ and $$\mathrm{var}(T)=\sum_jp_j\sigma_j^2+\sum_jp_j\mu_j^2-\left(\sum_jp_j\mu_j\right)^2$$