Для каждого значения параметра а решите неравенство 4^x-(2a+1)2^x+a^2+a<0

$2^{2x}-(2a+1) \cdot 2^x+a^2+a\ \textless \ 0 \\ \\ t=2^x \ (t\ \textgreater \ 0) \\ \\ t^2-(2a+1)t+a^2+a\ \textless \ 0 \\ \\ t_{1,2} =\frac{2a+1 \pm \sqrt{(2a+1)^2 - 4(a^2+a)}}{2}=\frac{2a+1 \pm \sqrt{4a^2 +4a+1-4a^2-4a}}{2}=\frac{2a+1 \pm \sqrt{1}}{2} \\ \\ t_1=\frac{2a+2}{2}=a+1; \ \ t_2=\frac{2a}{2}=a \\ \\ 2^x=a+1; \ \ 2^x=a \\ \\ x=\log_2(a+1); \ \ x=\log_2a \\\\ a+1\ \textgreater \ 0 \ \Rightarrow \ a\ \textgreater \ -1; \ \ a\ \textgreater \ 0 \\ \\$

$1) \ -1\ \textless \ a \leq 0; \ \ \ \log_2(a+1)<0$

      - +
---------------*---------------->x
       log_2(a+1)

$\boxed{x \ \textless \ \log_2 (a+1) ; \ \ -1\ \textless \ a \leq 0}$

$2) \ a\ \textgreater \ 0; \ \ \ (x-\log_2a) \cdot (x-\log_2(a+1)<0$
    +    - +
----------*-----------------*------------->x
   log_2(a) log_2(a+1)

$\boxed{\log_2 a \ \textless \ x \ \textless \ \log_2 (a+1); \ \ a\ \textgreater \ 0}$