Conservation of energy follows from invariance under translation in time, not inversion. This symmetry states that no matter when you do your experiment, it will give the same results. All isolated systems obey this symmetry (and therefore conserve energy) and no violation of it has ever been detected. (Needless to say, it would be a huge event if it were.)
In classical physics, only continuous symmetries - that is, symmetries that can be continuously connected to the identity transformation - have a corresponding conservation law. Quantum physics does permit conservation laws for discrete symmetries but these laws are far harder to visualize.
An example of this is conservation of parity, $P$, which corresponds to invariance under inversion in space, and which gives the parity - even or odd - of wavefunctions. Temporal inversion, $T$, is even harder to turn into a physical quantity because it requires a full relativistic treatment in which time is a coordinate like space and not a parameter (as it is in non-relativistic quantum mechanics). A third discrete symmetry is charge conjugation, $C$, which exchanges particles for their antiparticles.
It turns out that any consistent field theory must be invariant under all three operations when taken together - i.e. under $CPT$. Thus violation of parity - an experiment and its mirror image behaving differently - is possible, for example, if it comes together with violation of $C$ - i.e. the mirror experiment behaves like the original one if it is made of antimatter -, as was discovered in the sixties.
Violations of $C$ and $P$ together have also been discovered in recent years, which means that in some situations violations of $T$ must occur. The recent $B$-meson experiments confirm this. Since the $T$ symmetry does not correspond to energy but to a far more abstract quantity (which is not conserved), this does not lead to a nonconservation of energy.
No, this situation does not break time reversal symmetry, because the Gaussian also spreads out backwards in time. Its width is at a minimum at $t = 0$, and it increases as $t$ either increases or decreases. There's no preferred time direction at all.
One can intuitively understand this by looking at the Fourier components. The time $t = 0$ is the unique time all of the components "line up" to produce a purely real Gaussian.
Moreover, given a wavepacket that is time-asymmetric (e.g. it only spreads out going forward in time), you can always take the spread-out state and apply time reversal by complex conjugating it. This yields a wavepacket that narrows in time. This confirms that quantum mechanics really doesn't have a built-in arrow of time.
Physically, such a reversed wavepacket might not be realistic, in the exact same way that seeing an object spontaneously jump up from the floor isn't. This is the issue of the arrow of time, but it's much deeper than just looking at wavepacket spreading. You can't derive the arrow of time with just one-particle quantum mechanics.
Best Answer
You have all the elements in your question, your difficulty is about what is meant by "time reversal symmetry". Time reversal symmetry holds if, when "playing backwards", the motion observed obeys the same law. With friction it is not the case : friction opposes movement, when playing backwards it (seemingly) promotes it.
You can also go to equations for this. Let's have a damped oscillator of mass $m$: $$ m \frac{\mathrm d^2x}{\mathrm dt^2} = - c \frac{\mathrm dx}{\mathrm dt} - k x $$ Now play backwards with $\tau=-t$, thus $\mathrm d\tau/\mathrm dt = -1$: $$ m \frac{\mathrm d^2x}{\mathrm d\tau^2} = c \frac{\mathrm dx}{\mathrm d\tau} - k x $$
So, the equation is not the same—unless $c= 0$ : the physics of the backward time phenomenon is not the same if there's friction—while it is the same when there is no friction.