That's not an over-flow, that's actually a carry. An overflow is when the number of bits required to store the resulting number is greater than the number of bits you have available (8).

third: A+B = 15; 3+F+1=13 => 135

4
I think you meant 135

16? That would be the correct answer.

For the last one, if you have the binary number simply use that to convert the number to octal/hex.

Here's a trick for converting between binary and hex:

Every hex digit can be represented with 4 binary digits. So you only need to memorize the 16 binary digit combinations (or count up/down in binary) to get the matching hex digit:

(I'm using the Java hex notation, but the numbers on the left are binary)

0000 = 0x0

0001 = 0x1

0010 = 0x2

0011 = 0x3

0100 = 0x4

0101 = 0x5

0110 = 0x6

0111 = 0x7

1000 = 0x8

1001 = 0x9

1010 = 0xA

1011 = 0xB

1100 = 0xC

1101 = 0xD

1110 = 0xE

1111 = 0xF

Now to use this trick, simply take your binary number and group it into 4's and use that table to match your digits.

So,

1000 1010 0100 0101 0101 0101 0011 1001 1111 = 0x8A45559F

Converting to octals has the same trick: use groupings of 3.

100 010 100 100 010 101 010 101 001 110 011 111 = 424425251637

8
And how about base 4? If you guessed groupings of 2, you'd be correct.

Going backwards is equally as easy.

Unfortunately, there's no easy way to convert from 2 to decimal (well, not that I know of). You must simply convert using the basic "algorithm" for base conversion.