Electronic – Are \$h_{ie}\$ and \$r_{\pi}\$ the same thing

For BJT's there is a PN junction between the base and emitter. The arrow indicates the order of the junction (base to emitter or emitter to base). An NPN has stacked N, P, and N doped channels. The PN junction (between base and emitter) goes from the center out. PNP likewise is the opposite.

Observations, not necessarily fact:

In a MOSFET, the body is often connected to the source. For an N-channel MOSFET, the source is N-doped and the body is P-doped, so the arrow points from the source to the body. Likewise, a P-channel MOSFET has the reverse condition. Interestingly, Wikipedia has symbols for "MOSFETS with no bulk/body" which have opposite arrow directions. I have no good explanation for why these are this way, though I suspect it might follow a similar pattern and the semiconductor topology is different from "traditional" MOSFET topologies.

Your symbols for b (FET) are JFET symbols. Here, the PN junction is between the gate and "body" (semiconductor section connecting the drain and source; I don't know what the correct for this part of a JFET is so I just called it the body because it takes the bulk volume of the JFET). For an N-channel, the gate is P-doped and the body is N-doped, so the arrow points from the gate in. The P-channel JFET is the opposite so the arrow points out of the gate.

I've never used unijunction transistors (case d), but looking at the wikipedia page shows a similar doping structure as the JFET, the only difference the lack of an insulated gate (names also have changed, apparently it follows the "BJT" type naming of base and emitter). I would not be surprised if the arrow direction convention follows the order of the PN junction (wasn't immediately obvious to me which type the example structure on Wikipedia was for).

Additional info:

Bipolar Junction Transistors

MOSFET

JFET

unijunction transistors

Just to add some numbers: I simulated your circuit with the values given in your circuit and some sensible defaults (e.g. Vcc=9V and Vin=1V peak-to-peak).

Results are, that you operate the transistor very close to the maximum ratings.

I see a peak power dissipation of 627mW and an average power dissipation around 450mW. The maximum rating for the 2N3904 is 625mW, so no surprise that the transistor gets hot fast.

If you drop a TIP41C into the circuit you should still be able to measure a signal at the amplifier output, but it will probably a lot lower because the current gain of the TIP41C is a magnitude lower compared to the 2N3904.

As others already suggested you can run the TIP41 and the 2N3904 in darlington configuration. That'll will give you best of both worlds.

Nonetheless, from the learning experience I think your circuit is quite a success. It's running hard at the limit of the 2N3904 but otherwise it looks fine. I suggest that you take a look at differential amplifiers next. They are a very important building blocks for audio amplifiers.