Will this transistor setup work for powering a LED?
The T1 circuit will work as expected provided that the transistor supply voltage is &lt;= the base drive voltage. It is in your case so your circuit should be fine.

Here is an explanation of the trap that many fall into:



Figure 1. The "high-side" switch fail. Source.


If If  \$ V_{SS}&gt;V_{MICRO} \$ then the output protection diodes built into the chip provide a sneak path for the base current. 
The e-b junction of Q1 will be forward biased and current will flow through it, R1 and D1 to the micro supply. 
Q1 will turn on and LED will light and will not switch off.


Because your micro supply is 4.2 V and the transistor's emitter is at 3.3 V a high on the output will turn the transistor hard off.



LED current



Figure 2. Resistance calculation. Source.

Let's select a current of 5 mA for your LED. This is enough for most modern small LEDs to glow brightly. From Figure 2 we can see that at 5 mA an orange LED will drop about 1.65 V. Allowing about 0.3 V drop across your transistor that leaves 3 - 1.65 = 1.35 V across the current limiting resistor. From Ohm's law you can calculate the required value as \$ R = \frac {V}{I} = \frac {1.35}{5m} = 270 \ \Omega \$.

Base resistor


The base resistor we can use the rough approximation of \$ I_c = h_{fe}I_b \$ where \$ I_c \$ is the collector current, \$ h_{fe} \$ is the transistor gain and \$ I_b \$ is the base current.
The collector current is 5 mA and we can assume a gain of > 100 for a small signal transistor.
From the above \$ I_b = \frac {I_c}{h_{fe}} = \frac {5m}{100} = 0.05\ \mathrm{mA} \$.
A suitable base resistor is then \$ R = \frac {V}{I_b} = \frac {3}{0.05m} = 60 \ k\Omega \$. To be sure to drive the transistor fully into saturation most of us would pick a lower value to increase the base current. The transistor won't mind. 33 kΩ would be fine.