Finding solutions set of such equations in $\mathbb Z$, $\mathbb R$ etc Kindly solve for obtaining  solutions of the following equations in $\mathbb Z$, $\mathbb R$ etc
1) Solve in $\mathbb Z$ for $\frac{3}{\sqrt{x}} + \frac{2}{\sqrt{y}} = \frac{1}{\sqrt{2}}$
2) For M = { $(a-b)^2$ + $b^2$|a, b in $\mathbb Z$}. Prove that $2012$ not in $M$. Also prove that, 
     $M$ is a closed subset of $\mathbb N$ in respect of the multiplication of integers.
3) Solve in $\mathbb R$ for $5$$x^3$$ - 18$$x^2$$ + 43x - 6 = 12($$2^x$$)$
Thanks in advance.

We give rather full information towards 2), and much less full information towards 1). Question 3), which mixes polynomial and exponential, has less structure.  If you are looking for non-negative integer solutions $x$, you can just compute, trying $x=0$, $1$, $2$, and so on. You will find some solutions. And if you go far enough in your computations, you can prove that you got them all, by showing that after the numbers you have tested, $(12)2^x$ is bigger than the left-hand side. 
For closure, note the following famous identity, which can be verified by expanding each side.
$$(s^2+t^2)(u^2+v^2)=(su+tv)^2+(sv-tu)^2.$$
We show that $2012$ is not a sum of two squares, so in particular is not of the form described. The number $2012$ is equal to $4(503)$. It is easy to verify that if this is a sum of $2$ squares, say $2012=x^2+y^2$, then $x$ and $y$ must be even. So $503$ would be a sum of $2$ squares. We show that cannot happen. 
There are very few candidate squares, so we could go through them all, and check that none of them work.  Or else note that $503$ is congruent to $-1$ modulo $4$. It is easy to verify that a sum of $2$ squares cannot be congruent to $-1$ modulo $4$. For the square of an odd number is congruent to $1$ modulo $4$, indeed modulo $8$. 
For a), there are two things to do, one of which takes a fair bit of work. It is plausible, and should be proved, that our numbers $x$ and $y$ are of the form $2a^2$ and $2b^2$ respectively.
Assuming that, we are looking for positive integers $a$ and $b$ such that $\frac{3}{a}+ \frac{2}{b}=1$. A quick search will find them all, since $a$ and $b$ cannot both be big. Alternately, rewrite the equation as $ab-2a-3b=0$, and either solve for $a$, or rewrite as $(a-3)(b-2)=6$. 
The part about proving that solutions $x$ and $y$ must be of the shape we have described is a somewhat complicated variant of the proof that $\sqrt{2}$ is irrational. As a start, note that by squaring both sides we can conclude that $xy$ is a perfect square,
