Are all of the ARM opcodes 1 byte?

Question 1

No. There are now a few arm instruction sets. The primary arm instruction set, all instructions are 32 bits and the decoding of the instruction is not isolated to one field within the instruction. The thumb instruction set is based on 16 bit instructions, same answer though, depending on the instruction the decoding is in different places. Then there are thumb2 extensions to the thumb instruction set, same answer. The thumb2 instructions are one of the undefined instructions from thumb then add some more bits to distinguish which instruction.

Get one of the ARM Architecture Reference Manuals from infocenter.arm.com to see all of this. There are a couple of nice diagrams which show how this all works, the msbits are used to divide the instructions into different types then depending on the type the rest of the bits are decoded.

x86 comes from the 8 bit world where the designs tended to use one memory location(/access) (at the time that was a byte) for the more commonly used instructions, then have some opcodes multiplex into a second byte. Also immediates were added in their entirety. It was okay then, but is inefficient today. x86 should be the last instruction set you learn if ever, not representative of a good instruction set, you will need to unlearn a number of things to move forward.

wikipedia has a number of instruction sets on the wiki page itself, others it often has a direct link. IP vendors like arm and mips, as well as many chip vendors you can go right to their site and get the documentation.

Question 2

ARM instructions don't really have a concept of 'opcodes'. I don't know the x86 instruction set very well, but what you describe sounds like 6502 machine code where some bytes identify which instruction to execute and others specify data that the instruction uses.

Ignoring Thumb, all ARM instructions are four bytes long. The operands used by the instruction are contained within those four bytes.