why is \$P^{max}_{R3}=\frac{V^2_{th}}{4R_{th}}\$,not \$P^{max}_{R3}=\frac{V^2_{th}}{R_{th}}\$,why should we have to multiply \$R_{th}\$ with \$4\$?
Ignoring your error in calculating Vth, here's why the power is divided by 4.

Vth is the voltage at R4 with R3 disconnected, ie. the 'source' without a 'load'. In this case it is:-







simulate this circuit &ndash; Schematic created using CircuitLab

Now we attach a 'load' with resistance equal to Vth, and what happens?







simulate this circuit

The output voltage halves. 

\$P = \frac{V^2}{R}\$. With half the voltage \$P_L = P_{th} = \frac{(V_{th}/2)^2}{R_{th}}\$. 

\$(1/2)^2 = 1/4\$, so moving it to the bottom of the equation we get \$P = \frac{V_{th}^2}{4R_{th}}\$ 

In words:-
 The voltage is divided by 2 so the power is divided by 4 (compared to what it would be if Rth was 0 &ohm;).