Suppose $(a,b)=1$, then $(2a+b,a+2b)=1\text{ or }3$. Suppose $(a,b)=1$.  Let $d=(2a+b,a+2b)$.  
Then $d=(2a+b)u+(a+2b)v=a(2u+v)+b(2v+u)$ where $u,v \in \mathbb{Z}$.  
Since $(a,b)=1$, then $a(2u+v)+b(2v+u)=1$.
I'm not sure if I'm going in the right direction to solve this problem.
 A: Hint: 
Let $d=(a+2b,2a+b)$, so $d|3a$ since $d|2(2a+b)-(a+2b)$ and 
$d|3b$ 
since $d|2(a+2b)-(2a+b)$.
Therefore $d|(3a,3b)$.
A: Hint $\ $ Specialize the method below:  $\rm\,(x,y)\mapsto (2x+y,x+2y) =: (X,Y)\,$ has   $\, \det = 3,\,$ i.e.
$\qquad\begin{pmatrix} 2 & \rm 1 \\\\ \rm 1 & \rm 2 \end{pmatrix}\ \begin{pmatrix} x \\\\ \rm y \end{pmatrix}\ =\ \begin{pmatrix} X \\\\ \rm Y\end{pmatrix}\ \ \ \Rightarrow\ \ \ \begin{array} \rm\ 3\ x\ \ \ =\ \ \ \rm 2\ X \ - \ Y \\\\ \rm 3\ y\ =\ \rm\  - X + 2\ Y \end{array} $
Theorem  $\ $ If $\rm\,(x,y)\overset{A}\mapsto (X,Y)\,$ is linear then $\: \rm\gcd(x,y)\mid \gcd(X,Y)\mid \Delta \gcd(x,y),\,\ \Delta = \det A$
Proof $\ $ Inverting the linear map $\rm\,A\,$ by Cramer's Rule (multiplying by the adjugate) yields 
$$\rm \begin{eqnarray} a\ x\, +\, b\ y &=&\rm X\\ \\ \rm c\ x\, +\, d\ y &\ =\ &\rm Y\end{eqnarray}
\quad\Rightarrow\quad \begin{array} \rm\Delta\ x\ \ \ =\ \ \ \rm d\ X\, -\, b\ Y \\\\ \rm\Delta\ y\ =\ \rm -c\ X\, +\, a\ Y \end{array}\ ,\quad\ \Delta\ =\ ad-bc\qquad $$
Hence, by RHS system, $\rm\ n\ |\ X,Y\ \Rightarrow\ n\ |\ \Delta\:x,\:\Delta\:y\ \Rightarrow\ n\ |\ gcd(\Delta\:x,\Delta\:y)\ =\ \Delta\ gcd(x,y)\:.$
In particular $\rm\ n = \gcd(X,Y) \mid \Delta\, \gcd(x,y).\ $ 
Further, by LHS system $\rm\,n\mid x,y\ \Rightarrow\ n\mid X,Y\ \Rightarrow\ n\mid\gcd(X,Y).$
In particular $\rm\ n = gcd(x,y)\mid \gcd(X,Y).\ \ \ $  QED
A: Direct manipulation of gcd
$\gcd(a+2b,2a+b) = \gcd(a+2b,(2a+b)-2(a+2b)) = \gcd(a+2b,-3b)$
$ | \gcd(a+2b,3) \gcd(a+2b,b) | 3 \gcd((a+2b)-2(b),b) = 3 \gcd(a,b) = 3$
A: Let $$d=\gcd(a+2b, 2a+b).$$
$$\Rightarrow d|a+2b, d|2a+b.$$
$$\rightarrow d|2(a+2b), d| 2a+b.$$
Now use this result that if $d=\gcd(a, b), d|a, d|b\ \text{then}\ d|a\pm b$.
So,
$$d|2a+4b-2a-b.$$
$$\Rightarrow d|3b.$$
Also $$\Rightarrow d|3a.$$
$$\therefore d|(3a, 3b) \Rightarrow d= 1\ \text{or} \ 3. (\because (a, b)=1.)$$
