NPN transistor high resistance across collector and emitter
I am afraid there is a misunderstanding here.

you have killed the transistor by putting its base to 3.3V
you're measuring the resistance of the E-C which is pointless (and you may likely measure it with a multimeter with wrong polarity

This is the standard way of driving an NPN transistor:

(source: elinux.org)
Note that:

you always need a series resistor for the base. A transistor base will be and stay at 0,7V over the emitter (so in this circuit, at 0.7V) when the transistor conducts. If you force it to anything higher, then you killed the transistor. The other side of the series resistor is what you can drive with 3V or 5V. The resistance of the series resistor is not critical, anything around 1k-10k will do

the transistor will conduct in one direction (therefore using a multimeter and measure it in the wrong direction can give misleading results). That is, it is switching the GND to the load as seen on the schematic.
So instead of measuring resistance, you can measure voltage between 5V and the collector pin.


In the real life, the transistor &quot;amplifies&quot; current. I.e. when you put 3.3V to the resistor, there is a small current flowing from that to the base, and from the base to the emitter. The transistor &quot;amplifies&quot; this by its amplification factor (some datasheets shows as beta) which is around 200..500.
The amplification is in fact a trick, it is not generating bigger current for real. Rather, it can change the current flowing via the load, i.e. the current between collector and emitter. If you're using the transistor as a digital switch, a very small base current (say, 1mA) is enough to make C-E conduct and let up to 200-500mA (see above the apmlification factor) flow between the load (and C-E).
Transistor amplifiers use this principle to drive a power hungry speaker from a very small signal (however transistor power amplifiers are a more complicated topic).