During dead time of any DC-DC converter both the switches are turned off to avoid any shoot-through current. But in a synchronous Buck converter that would mean sudden zero current through inductor. Wouldn't that cause any problem?
There is no problem since during the dead-time the current will flow trough the body diode of the FET discharging the inductor, very similar as a non synchronous buck. After the dead-time the FET is turned on bypassing the conducting diode. With this action you can reduce the losses since the voltage across the FET will be much lower than the forward voltage of the diode.
All in all, Synchronous Buck is all about reducing the forward losses on the Buck diode. There is no change on the operation states of the converter itself. It will work in CCM, BCM and DCM given that you have the right dead-time.
Any inductor current is carried somewhere, for a while. There are various mechanisms to avoid this becoming a problem.
In an asynchronous buck with a switch+diode (I know you asked about synchronous, but let's start with one that you think should be OK, and show that it is less OK than you thought), the freewheel diode picks up the current, eventually. It can't do that instantaneously, as the inductance of the finite-length wires between the switch device and it is non-zero. This current first goes to charge 'unwanted' capacitances in the devices. In the event that the voltage on the switching transistor rises above its breakdown voltage, it will avalanche. Most switching devices have a 'maximum repetitive avalanche energy' specification, to cope with just this problem. Eventually, the high voltage across the wiring inductance causes the current to slew from the switch to the diode, and peace is restored.
In a high current design, great care is taken to minimise the switch/diode transfer inductance. This doesn't necessarily mean the inductance to each individually needs to be low, but it does mean they should be well coupled, so fat and close together.
How fast the current has to transfer is governed by the fall time of the switching FET. Adding a sniff of series resistance (a few ohms to the low 10s of ohms) is often used between gate driver and gate to increase the switching time a little. This is especially effective during switching as it limits the rate that the Miller capacitance charges, thus controlling the rate of voltage rise on the drain. While this increases dissipation in the off-going channel, it reduces transient overshoot which may improve EMI and other general behaviour.
So now you know what can happen in a an asynch converter, dead time does not seem so scary. First, the voltage rises on the off-going transistor, and it may avalanche. Second, current gets transferred into the rectifier device. Initially it's off, so it's first carried by the body diode. Eventually the channel comes on, which then takes up conduction.
If the wiring inductances or the current is so large that the switching transient exceeds the switch device's safe avalanche specification, then it's possible to use snubber components, typically a capacitor and resistor in series. These are best avoided if possible as they waste energy.
No, because you must start the dead-time at the time where the inductor current becomes zero. This means the inductor is discharged. This means you loose no energy.
You cannot suddenly make the inductor current zero, the inductor will resist that by making the voltage such that the current will flow anyway (and your transistors might be damaged then).