The concepts of "even" and "periodic" relate to the action of the one-dimensional analogs of wallpaper groups. "Even" means that the function is invariant under reflection about the origin, $t \to -t$, and "periodic" means it is invariant under translation by some period $p$, $t \to t+p$.
The function $f$ in the question has been declared over a fundamental domain, in this case the interval from $0$ to $\pi$. To extend it to a function of the reals, for each $t$ not in this domain we need to find a group element $g$ (generated by the reflection and translation) for which the application of $g$ to $t$ yields a number $t^{(g)}$ in the fundamental domain. We then define $f(t)$ to equal $f(t^{(g)})$.
Although this may sound abstract, it translates to a highly general, efficient, almost mindless computational method. For one-dimensional groups, we can figure out $t^{(g)}$ by means of the Mod
function (to implement the translation) and the Abs
function (to implement the reflection). It is convenient first to translate $t$ into a domain that is as close to the origin as possible: do this by offsetting Mod
by half the period, as in Mod[t, 2 Pi, -Pi]
. The result now lies in the symmetrical interval $[-\pi, \pi]$ around $0$. Taking the absolute value assures the result is positive, where $f$ can be applied. Whence, after defining f
as in the question, merely declare its extension g
by following this recipe:
f[t_] := Piecewise[{{3 t, 0 <= t <= Pi/2}, {3 t + 6, Pi/2 <= t < Pi}}, Null];
g[t_] := f[Abs[Mod[t, 2 Pi, -Pi]]];
Plot[{g[t], f[t]}, {t, -10, 10}, PlotStyle -> {Thick, Thick}]

Visually, it is clear that $g$ (the blue graph) is even and periodic and agrees with $f$ (the red graph) on the fundamental domain $[0, \pi)$ where $f$ is defined.