Calculation of overall rate constant of a parallel reaction
The key to answering these kinds of riddles is to think about units.  Rates of reaction have units of $\mathrm{\frac{mol}{L\cdot s}}$, but that' isn't fully specified.  You need to know moles of what.  

In your case, the confusion is stemming from confusion of what the "moles" in $\mathrm{\frac{mol}{L\cdot s}}$ refers to.

The answer can either be moles of "reaction", or moles of A.  The other key thing to realize is that you can write "the" reaction however you want.  The value of the "rate constant" $k$ depends on how you define the rate.  If for the overall reaction you write:

$$\ce{3A -&gt; \varnothing}$$

...and if the "rate" you are calculating is the rate of reaction, then there are three moles of A lost per mole of reaction, so 

$${dA \over dt}=-3k_{\alpha}\ce{[A]}$$

If instead you decide "the" reaction is 

$$\ce{A -&gt; \varnothing}$$

then you can rate the rate of disappearance of A as 

$${dA \over dt}=-k_{\beta}\ce{[A]}$$

The choice is up to you and it is arbitrary.  You just have to keep track of the moles of what the rate laws you write refer to.


  Then to calculate the k where k is the rate constant for the overall reaction we write [[Equation (i)]]


You need to think carefully about what you meant by the words "the overall reaction".  When you write equation 1, you are adding together rates of loss of A, i.e. the units are $\mathrm{\frac{mol~of~A}{L\cdot s}}$.  The only reaction you can write that has the same rate is one where the coefficient of A is 1.

"The" half-life for radioactive elements usually refers to units where units are $\mathrm{\frac{mol~of~element}{s}}$.  The fact that the unit is $\mathrm{\frac{mol~of~element}{s}}$ and not $\mathrm{\frac{mol~of~reaction}{s}}$ means that if we wanted to write the corresponding chemical reaction, the stoichiometric coefficient had better be 1.$%edit$