For (a), you don't need to complete the square. If $a,b,c$ are consecutive then they must be of the form

They must satisfy Pythagoras' theorem, which should enable you to find $a$.

For (b), the condition implies that

1. Suppose $a,a+1,a+2\in \mathbb{N}$ form a Pythagorean triple. Then

   $$\begin{array}{rl}(a+2{)}^2& =(a+1{)}^2+a^2\\ a^2+4a+4& =a^2+2a+1+a^2\\ a^2-2a-3& =0\\ (a-3)(a+1)& =0\\ a& \in \{-1,3\}\end{array}$$

   So $a=3$ is the only solution.

2. Suppose $a,b,c\in \mathbb{N}$ is a Pythagorean triple, and that

   $$\begin{array}{rlrl}& & c-b& =2(b-a)\\ \Rightarrow & & c& =3b-2a\end{array}$$

   Then, since $a,b,c$ is Pythagorean, we have

   $$\begin{array}{rl}{c}^2& =b^2+a^2\\ (3b-2a{)}^2& =b^2+a^2\\ 9b^2-12ab+4a^2& =b^2+a^2\\ 3b^2-12ab+3a^2& =0\\ b^2-4ab+a^2& =0\\ (b-2a{)}^2-4a^2+a^2& =0\\ (b-2a{)}^2& =3a^2\\ b-2a& =\pm \sqrt{3}a\\ b& =2a\pm \sqrt{3}a\\ b& =(2\pm \sqrt{3})a\end{array}$$

   Now if $a\in \mathbb{N}$, we can't possibly also have $b\in \mathbb{N}$ due to the $\sqrt{3}$ on the right-hand side.

   This is a contradiction, so we must conclude that such a Pythagorean triple does not exist.