For elliptic curve $y^{2}=x(x+a^{2})(x+(1a)^{2})$,($a$ is a rational number and does not equal 0,1,1/2),is its rank always 0?

Your family of elliptic curves are exactly those elliptic curves with torsion subgroup containing $\Bbb Z / 2 \Bbb Z \times \Bbb Z / 4 \Bbb Z$. To see this make the linear change of variables $t = a/(1a)$ and look at the parameterizations on page 101 of Husemöller's book "Elliptic Curves". So in particular there are values of $a$ for which your elliptic curve has MordellWeil rank 8, the first of which was discovered by Elkies in 2005. See Dujella's website. Moreover, Eroshkin found that there are infinitely many such curves with rank at least 5. These happen to be my favorite elliptic curves. I'd be very interested to know why you were interested in them. 


Although Junkie has answered the question, I'd like to point out that in the case of parametrized families of elliptic curves (such as this) it is often easy to find an explicit subfamily with positive rank. In the present case, let us take $x=2a^2$ and see what condition on $a$ forces this to give a point on the elliptic curve. Making this substitution we obtain. $$ y^2=6 a^4 (3 a^22a+1). $$ We can simplify by defining $b=y/6 a^2$. Thus $$ 6b^2=3 a^22a+1. $$ This is a conic with the point $(a,b)=(1,1)$ and so we can parametrize all the solutions: $$ a=(t^2 + 4t  10/3)/(t^2  2), \qquad b=(t^2  8t/3 + 2)/(t^2  2) $$ where $t$ is rational. The argument gives that $$ (x,y)=(2a^2,6a^2 b)$$ is a rational point on the elliptic curve provided $a$, $b$ have the above shape. It should be possible through a slightly tedious computer algebra calculation to determine all rational numbers $t$ where the point above is torsion (using Mazur's Theorem), and for all other values of $t$ the rank is positive. 


I'm probably going out on a limb here, but I think that the following conjecture is reasonable. Conjecture. Let $E:y^2=x^3+a(T)x+b(T)$ be an elliptic curve defined over $\mathbb{Q}(T)$ with the property that its $j$invariant is not in $\mathbb{Q}$. Then there are infinitely many rational numbers $t$ such that $E_t(\mathbb{Q})$ has positive rank. Of course, I have no idea how one might prove this. As Samir pointed out, in some cases one can find a covering $C\to \mathbb{P}^1$ such that the pullback of $E$ has positive rank over $C$ and such that $C$ has genus 0 (or 1) with infinitely many rational points. But in general this won't be possible. One final note. It's possible to create families for which the sign of the functional equation is always even, in which case the fibers have even rank (if we believe BirchSwinnertonDyer). But even in this case, I'd expect infinitely many fibers to have rank 2 (or greater). 

