This is the sort of question you are likely to stumble across in an IB HL Paper 2 examination. It would probably carry a few more marks and thus require a longer answer, but that is reason to panic!
First of all, let's define what SN1 and SN2 mean. The 'S' component means the reaction is a substitution reaction, i.e. an atom or group on the molecule is being substituted (swapped) for another. The 'N' component (normally subscript) means that it is nucleophilic, i.e. the reaction involves a nucleophile. A nucleophile is a species that 'is attracted to a nucleus', in this case a partial positive charge on an atom.
Before we go any further, let's pick an exemplar reaction. Let us take 1-bromobutane (commonly known as butylbromide). We are going to react it with a dilute solution of sodium hydroxide. In solution, sodium hydroxide (NaOH) will produce hydroxide ions, :OH-. This species is an excellent nucleophile as the oxygen atom has a lone pair of electrons, which can attack the electrophile (species that likes electrons), which in this case, is the carbon atom bonded to the bromine atom in our 1-bromobutane.
The carbon atom is electron deficient, as the bromine atom is electronegative. This means it pulls the electron density of the covalent bond towards itself, giving it a partial negative charge. The carbon atom is thus left electron deficient, and so has a partial positive charge.
Let's jump ahead a bit and look at the product(s) that we obtain from this reaction. We would get butan-1-ol and bromide ions in solution. How did we get this, and which mechanism should we use?
We've clearly substituted a bromine atom for a hydroxyl group, so we know that it's a substitution reaction, and as hydroxide is a nucleophile, we know it's nucleophilic. But is it SN1 or SN2?
We know from experiments that this particular reaction proceeds via SN2. We know this by looking at the reaction kinetics, or the rate of reaction. The rate equation for this reaction is:
rate = k[CH3CH2CH2CH2Br] [-OH].
The square brackets indicate concentrations. This rate equation means that the reaction rate depends on both the concentration of our halogenoalkane and our nucleophile, so the rate determining step involves both species. The hydroxide attacks the carbon atom at the same time as the bromine atom leaves to form bromide ions. This is thus an SN2 reaction - the '2' denotes that two species are involved in the rate determining step.
What if we changed our electrophile (halogenoalkane) to 2-bromo-2-methylpropane? This molecule would proceed via SN1 mechanism. This means only one species is involved in the rate determining step, so the rate equation would look like:
rate = k[C(CH3)3Br]
No matter how much hydroxide you have present, it makes no difference to how fast the reaction goes. In this case, in the first step, the bromine atom leaves as bromide to form a carbocation intermediate. This means that the carbon once bonded to bromine now has a formal positive charge. Once this slow step has finished, the hydroxide ion can now attack to form 2-methylpropan-2-ol.
We have now answered (very thoroughly!) the first part of the question. In the exam, you would summarise what we've discussed above into a few sentences. Chemists detest writing, we like to keep things short and to the point!
How do we know then via which mechanism the reaction will go? There are two main factors - sterics and electronics.
Let's look at sterics first. 1-bromobutane is a primary halogenoalkane. This means the central carbon atom has only one alkyl group bonded to it, in this case a propyl group. The theoretical transition state in an SN2 reaction has a carbon atom with 5 groups bonded to it - two hydrogens, the alkyl group, the leaving bromine atom and the attacking hydroxyl group. That's a lot for one atom to hold, but it can manage as hydrogen atoms are tiny. In the case of 2-bromo-2-methylpentane, if the reaction went by SN2, the theoretical transition state would again have 5 groups bonded to the central carbon atom, but in this instance the small hydrogens are replaced by much bigger, bulkier methyl groups. It can't all fit in space, so this mechanism isn't really viable.
Why then don't primary halogenoalkanes react via SN1? You now know that SN1 reactions go via a carbocation intermediate. Carbocations are very unstable - nature hates charged species! In the case of a tertiary halogenoalkane (3 alkyl groups bonded to central C atom), this formal positive charge is stabilised by the positive inductive effect of the neighbouring methyl groups. Alkyl groups push electron density towards the electron deficient carbon atom, making it more stable. For a primary halogenoalkane, there is only one alkyl group able to do this - a primary carbocation is far too unstable and so never forms. These are known as electronic effects.
You may notice I've left out secondary halogenoalkanes. This is because, in all honesty, chemists can rarely predict which mechanism they will use, so best stick to primary and tertiary.
You should now be able to define SN1 and SN2 reactions, and understand why molecules react differently depending on their characteristics. Good luck!