A continuous function defined on an interval can have a mean value. What about a median? A function can have an average value
$$\frac{1}{b-a}\int_{a}^{b} f(x)dx$$
Can a continuous function have a median?
How would that be computed?

Let $X$ be a random variable uniformly distributed over the interval $[a,b]$.  If $f : [a,b] \to \mathbb R$ is a (Borel) measurable function, then $Y = f(X)$ is also a random variable, and its distribution can be characterized by all the usual statistical properties, such as the mean and the median.
In particular, the mean of the function $f$, as defined in your question, corresponds exactly to the mean of the random variable $Y$, as given by the law of the unconscious statistician.
Similarly, we can naturally define the median of the function $f$ to be the median of the corresponding random variable $Y$, i.e. a value $m$ such that $\mathrm{Pr}(Y \ge m) \ge \frac12$ and $\mathrm{Pr}(Y \le m) \ge \frac12$.
In particular, since $\mathrm{Pr}(Y \ge m) = \frac{1}{b-a}\mu(\{x \in [a,b]: f(x) \ge m\})$, where $\mu(S) = \int_S 1\, dx$ denotes the measure of the set $S$, and conversely for $\mathrm{Pr}(Y \le m)$, it follows that the median $m$ of $Y$ (and thus of $f$) is simply the value such that the graph of $f$ lies at or below $m$ for half of the interval $[a,b]$, and at or above $m$ for the other half.
(The median, as defined above, may not be unique if $f$ is not continuous. However, if there are multiple values of $m$ that satisfy the definition above, then they will form a single interval.  In this sense, the situation is no different from that of the median of a probability distribution in general.)
