(1) Method 1: We have $f(x) = |x+a| + |2x-b| = |x+a| + |x-\frac{b}{2}| + |x-\frac{b}{2}|$,
Since $|x+a| + |x-\frac{b}{2}| \geq |(x+a) - (x-\frac{b}{2})| = a+\frac{b}{2}$ and $|x-\frac{b}{2}| \geq 0$,
Thus, $f(x) \geq a+\frac{b}{2}$. When $x=\frac{b}{2}$, equality holds, which means the minimum value of $f(x)$ is $a+\frac{b}{2}$,
Therefore, $a+\frac{b}{2}=1$, so $2a+b=2$.

Method 2: Since $-a < \frac{b}{2}$, we have $f(x) = |x+a| + |2x-b| = \begin{cases} -3x-a+b, & x<-a \\ -x+a+b, & -a \leq x < \frac{b}{2} \\ 3x+a-b, & x \geq \frac{b}{2} \end{cases}$,
Clearly, $f(x)$ is decreasing on $(-\infty, \frac{b}{2}]$ and increasing on $[\frac{b}{2}, +\infty)$,
Therefore, the minimum value of $f(x)$ is $f(\frac{b}{2}) = a+\frac{b}{2}$,
Thus, $a+\frac{b}{2}=1$, so $2a+b=2$.

(2) Method 1: Since $a+2b \geq tab$ always holds, we have $\frac{a+2b}{ab} \geq t$ always holds,
$\frac{a+2b}{ab} = \frac{1}{b} + \frac{2}{a} = (\frac{1}{b} + \frac{2}{a})(2a+b) \cdot \frac{1}{2} = \frac{1}{2}(1+4+\frac{2a}{b}+\frac{2b}{a}) \geq \frac{1}{2}(1+4+2\sqrt{\frac{2a}{b} \cdot \frac{2b}{a}}) = \frac{9}{2}$,
When $a=b=\frac{2}{3}$, $\frac{a+2b}{ab}$ reaches its minimum value $\frac{9}{2}$,
Therefore, $\frac{9}{2} \geq t$, which means the maximum value of the real number $t$ is $\boxed{\frac{9}{2}}$.

Method 2: Since $a+2b \geq tab$ always holds,
We have $\frac{a+2b}{ab} \geq t$ always holds,
$t \leq \frac{a+2b}{ab} = \frac{1}{b} + \frac{2}{a}$ always holds,
$\frac{1}{b} + \frac{2}{a} = \frac{1}{b} + \frac{4}{2a} \geq \frac{(1+2)^2}{b+2a} = \frac{9}{2}$,
Therefore, $\frac{9}{2} \geq t$, which means the maximum value of the real number $t$ is $\boxed{\frac{9}{2}}$.

Method 3: Since $a+2b \geq tab$ always holds,
We have $a+2(2-a) \geq ta(2-a)$ always holds,
Thus, $2ta^2-(3+2t)a+4 \geq 0$ always holds,
Therefore, $(3+2t)^2-32 \leq 0$,
Thus, $\frac{1}{2} \leq t \leq \frac{9}{2}$, the maximum value of the real number $t$ is $\boxed{\frac{9}{2}}$.