To check the solution via radicals Given equations are 
$$x^5+y^5=a \tag 1$$
$$5xy(x^2+xy+y^2)=b \tag 2$$
Is it possible to find   $x$ or $y$ via using radicals?

My attempt
$$x^2+y^2=\frac{b}{5xy}-xy$$
$$x^4+y^4=\frac{b^2}{5^2x^2y^2}-x^2y^2-\frac{2b}{5}$$
$$(x^2+y^2)(x^4+y^4)=\left(\frac{b}{5xy}-xy\right)\left(\frac{b^2}{5^2x^2y^2}-x^2y^2-\frac{2b}{5}\right)$$
$$x^6+y^6+x^2y^2(x^2+y^2)=\left(\frac{b}{5xy}-xy\right)\left(\frac{b^2}{5^2x^2y^2}-x^2y^2-\frac{2b}{5}\right)$$
$$x^6+y^6+x^2y^2\left(\frac{b}{5xy}-xy\right)=\left(\frac{b}{5xy}-xy\right)\left(\frac{b^2}{5^2x^2y^2}-x^2y^2-\frac{2b}{5}\right)$$
$$x^6+y^6=\left(\frac{b}{5xy}-xy\right)\left(\frac{b^2}{5^2x^2y^2}-2x^2y^2-\frac{2b}{5}\right)$$
I haven't got $x^5+y^5$ in my way,it looks impossible to solve it via radicals but I need to proof.
Thanks for answers

No. First, notice that
$$ (x+y)^5 = (x^5+y^5) + 5 x y (x^2+xy+y^2) (x+y)$$
So your equations imply
$$(x+y)^5 = a + b(x+y)$$
Putting $s = x+y$, if $x$ and $y$ could be expressed in terms of $a$ and $b$ using radicals, then so could the root of $s^5 -bs -a$. This quintic cannot be solved in radicals. (In fact, this quintic is in what is known as Bring-Jerrard normal form. Every quintic can, by a change of variables be put into this form so, if we could solve equations in this form by radicals, we could also solve general equations.)

So, how did I find this? I always try to use symmetries of the equations to reduce their degree before starting. In this case, the only symmetry I noticed was switching $x$ and $y$. So I set $s=x+y$ and $t=xy$, and used the fundamental theorem of symmetric functions to write:
$$s^5 - 5 s^3 t + 5 s t^2 = a \quad 5 t (s^2-t) = b $$
The next thing I wanted to do was eliminate one of the variables. I wasn't smart enough to do it by hand, so I asked Mathematica:
(* Find an equation satisfied by t, containing no copies of s. *)
GroebnerBasis[{s^5 - 5 s^3 t + 5 s t^2 - a, 5 t (s^2 - t) - b}, {t}, {s}]

Mathematica output:
{b^5 - 25 b^4 t^2 + 125 b^3 t^4 - 3125 a^2 t^5 + 625 b^2 t^6 - 3125 b t^8 + 3125 t^10}

A degree $10$ equation. It probably isn't solvable by radicals, but it seems hard to prove. Let's try again:
(* This time, keep s and eliminate t *)
GroebnerBasis[{s^5 - 5 s^3 t + 5 s t^2 - a,  5 t (s^2 - t) - b}, {s}, {t}]

Output:
{-a - b s + s^5}

Ah. much simpler. And clearly not solvable by radicals, so neither is the original problem.
