$\angle DAC=\angle DAB=\angle ACB$ so $DA=DC$. From the similar triangles, we have $\frac{AD+DB}{BA}=\frac{CA+AB}{BC}\implies a^2=c(b+c)$. But since $\gcd(c,b+c)=1$, both $c,c+b$ must be squares, so we are done with part (a). If $c=n^2$, then $n\mid a$ so $\gcd(a,c)\geq n>1$, a contradiction. So there is no such triangle, I think.

You sure about part (b)? Try $a=12$ , $b=7$ and $c=9$. I think it works.

For a fixed $n$ in $b)$ we can take $(a,b,c) = (n^{2} + n, 2n+1, n^{2})$.