- Just show via an induction that $b_n\le b_{n+1} \le a_{n+1} \le a_n$: this proves that both sequences are convergent.
- Then take the limit in the definition and the previous inequality: you get
$$A = \frac 12 (A+B)
\\A\ge B$$so $A=B.$
details for 1.:
a) The inequality
$$
u<v\implies \frac {u+v}2<v
$$is trivial.
b)$$
u<v\implies \frac 1u > \frac 1v
\\ \implies \frac 1u > \frac 12 \left(\frac 1u +\frac 1v\right)
=\frac{u+v}{2uv}\implies u< \frac{2uv}{u+v}
$$
c) As $0\le(\sqrt{u}-\sqrt{v})^2$,
$$
\sqrt{uv}\le \frac{u+v}2\\
4uv\le (u+v)^2\\
\frac{2uv}{u+v} \le \frac {u+v}2
$$
One of the (probably, the) most important techniques for finding explicit formulae for sequences defined by recurrences is that of generating functions. (In this case it might be overkill but it is worth learning it at some point in my opinion.)
The idea is (1) to define the formal power series
$$
B(x)=\sum_{n\geq 1} b_{n} x^{n}, \qquad\qquad \mbox{(the "generating function")}
$$
(2) to translate the recurrence into some equation for $B(x)$, (3) to solve this equation for $B(x)$, and, finally, (4) to re-expand $B(x)$ to get the answer.
In this case, take the recurrence
$$
b_{n+1}=2b_n+2^{n-1},
$$
multiply it by $x^{n+1}$ and sum over $n\geq 1$ to obtain a relation about the generating function $B(x)$. Proceeding separately for each term in the recurrence, we have
\begin{align*}
\sum_{n\geq 1}b_{n+1}x^{n+1} &= b_2x^2+b_3x^3+\cdots=B(x)-b_1x=B(x),
\\
\sum_{n\geq 1}2b_{n}x^{n+1}&=2xB(x),
\\
\sum_{n\geq 1}2^{n-1}x^{n+1}&=\frac{x^2}{1-2x}.
\end{align*}
In the first equality we used that $b_1=0$ by assumption (but the method would work for any value of $b_1$), and in the third equality a geometric series.
Collecting these pieces, the recurrence for the coefficients is equivalent to the (algebraic, in this case) relation
$$
B(x)=2xB(x)+\frac{x^2}{1-2x}.
$$
We can solve for $B(x)$:
$$
B(x)=\frac{x^2}{(1-2x)^2}.
$$
The final step consists in re-expanding in Taylor series $B(x)$ to read off the coefficients. With some experience you recognize that here you need the expansion
$$
\frac y{(1-y)^2}=\sum_{n\geq 1}ny^n.
$$
(This is, essentially, a derivative of the geometric series $\frac 1{1-y}=\sum_{n\geq 0}y^n$.)
Apply it with $y=2x$ to get
$$
B(x)=\frac x2\frac {(2x)}{(1-2x)^2}=\frac x2\sum_{n\geq 1}n(2x)^n=
\sum_{n\geq 1}n2^{n-1}x^{n+1}=\sum_{n\geq 1}(n-1)2^{n-2}x^{n}
$$
finally yielding the desired answer
$$
b_n=(n-1)2^{n-2}.
$$
A good reference (but there are plenty) for this method is
Wilf: Generatingfunctionology (available at https://www2.math.upenn.edu/~wilf/DownldGF.html)
Best Answer
Hint To show thay are well defined, you need to check that $a_n+b_n$ can't be $0$ (for all $n \geq 1$), so $\left(b_n\right)$ exists.
For the limit, you might notice that $$ a_{n+1}b_{n+1}=\frac{a_n+b_n}{2}\times\frac{2a_nb_n}{a_n+b_n}=a_nb_n $$ (under condition you've proved that $a_n+b_n \ne 0$ for all $n \geq 1$). The you can deduce that
$$ a_{n+1}=\frac{1}{2}\left(a_n+\frac{a_1b_1}{a_n}\right) $$ And maybe you know this type of sequences ? This is a "Babylon Algorithm".