L1 is **not** the complement of {a^{n}b^{m} where n=m}. Because complement means (a+b)*-L1 = (a+b)*-a^{n}b^{n }which can generate many strings like {abab,babba ..} basically anything which are not of a^{n}b^{n }this form. However we can still create DPDA for language L1.

Please refer to the 2nd answer here https://www.quora.com/What-would-be-the-PDA-for-a-n-b-m-where-n-neq-m

[Don't see the other answers there]

So L1 is a DCFL.

Second one, that also can be reduced to the first problem. It says that the number of a's and b's are not equal (either n(a)>n(b) or n(a)<n(b) ). So ultimately same as above. And we don't need to take any action(pushing or popping) for c's as it doesn't need to be compared with anything. So this is also a DCFL.