Question related with the Calculation.(Easy but calculation process complicated.) Though idea for solving of these problems surely simple, I tried these over and over and Whenever I did it, My answer came out differently for each attempts. :( 
(The Calculation process might be little complicated.)
Please help me. 
$Q-1)$
$a+b = \sqrt {2 \sqrt 3 - \sqrt 2}, a-b= \sqrt {3\sqrt2 - \sqrt 3}$
Find the $a^4 + b^4 + 3a^2b^2$
$Q-2)$
Let the given right triangle that its perimeter's length is $2a$ and length of the Hypotenuse is  $x$
Then Find the range of the $x$ 
My attempt for the Q1
Since $(a+b)^2 = {2 \sqrt 3 - \sqrt 2}$ and $(a-b)^2 = {3\sqrt2 - \sqrt 3}$
$a^2 + b^2 = {1 \over 2}{(\sqrt3 + 2\sqrt 2)}, ab = {1 \over 4}({3 \sqrt3 - 4\sqrt 2})$
$(a^2 + b^2)^2 + a^2b^2  = {1 \over 16}{(103- 8\sqrt 6)}$ [$Not$ $10$] 
It seems the the answer is incorrect I thought for the Q1. But the one who suggest this question said the answer like the below.
The Answer) 
$Q1)$ $10$
$Q2)$ $2(\sqrt2 - 1)a \le x < a$
Thanks.

Your answer to part $1$ is correct.
For part $2$, let the interior angle between the leg and hypotenuse of the right triangle be $\theta$. Clearly, $0<\theta<\pi/2$. The perimeter of the triangle$$x+x\cos\theta+x\sin\theta=2a\implies x=\frac{2a}{1+\cos\theta+\sin\theta}$$For the given range of $\theta$, you have $1<\sin\theta+\cos\theta=\sqrt2\sin\left(\frac\pi4+\theta\right)\le\sqrt2$, so you get$$\frac{2a}{1+\sqrt2}\le x<\frac{2a}{1+1}$$
