the transistor for increasing current output of linear voltage regulator
First, to answer your question, you want a PNP power transistor rated for at least twice the peak input voltage and twice the max output current.  The transistor will see less than either of these max parameters, but that is design margin to increase reliability.  Since you are trying to make 0.9 A, go for a 2A to 5A rated transistor.  More is fine, but more expensive for no benefit.  You don't say what the input voltage is.  If it is 12 V, use a transistor rated for at least 25 V.  etc.  When the circuit is running, Q1 will see only 7 V, but again, design margin.
The value of R1 determines when Q1 starts to supply current to the output.  You can select this resistor such that when the circuit is running at its max output current, approx. 50% of the regulator circuit heat is dissipated in Q1 and 50% in U1 (the 7805).  If you wanted the circuit to supply more than U1 can handle on its own, such as 5 A, then you pick R1 such that the U1 current is about 75% of its rated max., and then Q1 would pass everything else and dissipate most of the heat.
Vbe is larger in power transistors than in small signal transistors, so let's assume that Q1 &quot;comes on&quot; at 0.7 V.  The current through R1 is the same as the current through U1.  At 0.45 A (1/2 the output current) through U1, you can solve for R1 with Ohm's Law.  For currents above 0.45 A, the difference will go through Q1.
To drill down further, the heat dissipated in Q1 and U1 will not be exactly equal because some power is dissipated in R1.  Also, the Q1 base current flows through U1, so it adds a little heat to both devices.  Neither of these is a big deal, just things that will make the actual results slightly different from calculations.