To be a stereocenter you need to:

• be a carbon
• be sp³ hybridized (all single bonds)
• have four different substituents

Are there other types of stereocenters? Of course! ( sp³ sulfur atoms, for example). But in undergraduate organic chemistry, 99% of all sterocenters you come across will be sp³ hybridized carbon atoms.

So in the molecule below, there are four stereocenters.

Note that the sp² (alkene) carbons can't be sterocenters.

This is two separate problems. Many textbooks describe how to mentally convert " zig-zag" (wedge/dash) structures to Fischer projections, but I've never met a student who can do this without making mistakes. So you should convert the name ((2R,3S) 2-bromo-3-chlorobutane) to a zig-zag structure, and then convert the name to a Fischer projection. Never try to convert a zig-zag structure directly to a Fischer projection.

To draw the zig-zag structure, first draw a structure with each halogen (the highest priority substituents) as a wedge, see the R/S configuration we drew, and then adjust as necessary. Why wedges? Because that puts the hydrogen as a dash, so R/S is easy to assign. With two wedges, the structure is (2R, 3R). So the 2R position is fine, and we just switch the wedge at C-3 to a dash, and the structure is correct (2R,3S).

To draw the Fischer projection, we do something similar- arbitaritly draw a structure, check R/S, and then adjust as necessary. In the structure below, I drew (arbitarily) 2S,3S, but we need 2R,3S, so I just switch C-2 and we have the correct structure.

The methyl group is bulky and so is most stable in the equatorial position.

The 1,4 cis compound's most stable chair form has its leaving group (Br) and beta-hydrogen in an anti-coplanar (anti-periplanar in some textbooks) conformation.

So most of the time, 1,4 cis is in a chair conformation that is able to undergo an E2 reaction.

The opposite is true for the 1,4 trans compound. Most of the time it is in a chair conformation that is unable to undergo an E2 reaction.

Therefore, the 1,4 cis compound will undergo an E2 elimination reaction faster than the 1,4 trans compound.

The reaciton of an alkene with Br₂ is an anti-addition, and hydrogenation (H₂ or D₂) is a syn addition. From this we can figure our the relative stereochemistry of each product.

Each starting material is achiral, and therefore not optically active, so the products cannot be optically active. 

There are three ways for products not to be optically active:

• products can be achiral
• products can be a racemic mixture
• products can be meso

In both a) and b) each product has a sterocenter, so the products can't be achiral.

In a), one stereocenter is S and the other stereocenter is R, and both with the same substituents, so this is a mirror image relationship, and the products are meso.

In b), the products have no internal mirror planes, so the products are chiral, and must be a racemic mixture.

The starting material (2-aminobutane) is a racemic; it has equal amounts of R and S enantiomers.

Enantiomers have the same the same chemical and physical properties, so we can't separate them using common lab techniques, such as recrystallization, chromatography, etc. But diastereomers can have different properties, and we can exploit this fact to separate otherwise inseparable compounds.

Adding an optically pure acid to the amine produces a diastereomeric mixture of salts which can be separated. Adding a base (NaOH) "breaks" the salt and allows us to isolate the pure R or S amine.

We are able to separate the (R,R) and (R,S) salts because they are diastereomers and so have different chemical and physical properties.

Notice that when drawing resonance forms with positive charges, the arrows never come from the positive charge. Arrows only come from π (pi) electrons- lone pairs or double/triple bonds.