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To take your first example, the statement "7 ≡ 2 (mod 5)" is equivalent to "7  2 is divisible by 5". So, you can equally well say that 17 ≡ 2 (mod 5), because 17  2 = 15 is divisible by 5. Also, 2 ≡ 3 (mod 5), because 2  (3) = 5 is divisible by 5. So, to address your final point, two numbers can be congruent to some modulus even if one is positive and one is negative, or if both are negative. Thus, 3 ≡ 8 (mod 5), because 3  (8) is divisible by 5.
In the first problem, to find c we first calculate 13a as an ordinary number: We are given a ≡ 11 (mod 19), so we can use a = 11. Thus, 13a = 13*11 = 143. Now we must find the integer c, with 0 ≤ c ≤ 18, such that c ≡ 143 (mod 19). Some calculators will do this for you  Google will calculate it if you search for 143 % 19. It gives the answer as 10. (Here, % is the modulo operator, not to be confused with the percent symbol!)
To work it out yourself, you could subtract multiples of 19 from 143 until you reach a number, c, in the range 0 ≤ c ≤ 18. That could take a while, though, especially with a much bigger number than 143. A better method is to divide 143 by 19, and take the remainder, which will always be in the range 0 ≤ c ≤ 18. If your calculator doesn't have a remainder button, just calculate 143/19, getting 7.526315... . This tells you that the quotient is 7, so to find the remainder, you just calculate 143  (7*19), getting c = 10, the same as Google. :)
Just to reiterate, "143 ≡ 10 (mod 19)" is equivalent to "143  10 is divisible by 19". The latter is true because 143  10 = 133 = 7*19.
In the second problem, the numbers are so small that you can calculate the answer most easily by just subtracting 23 a couple of times:
43  23 = 20.
20  23 = 3.
So 3 ≡ 43 (mod 23), and the answer is a = 3.
Again, this makes sense because 43  (3) = 46 is divisible by 23.
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