Here is a start:

Problem 1:

a+b+c = 11

a = b+c+1

a×100+b×10+c = c×100+b×10+a+396

The third one can be reduced to: a×100+c = c×100+a+396

And then to: a×99 = c×99+396

And then to: a=c+4

(Yet to be solved ... anyone?)

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Problem 2:

A bit tricky deciding what length is which. Working on the clues backwards:

The perimeter tells us: a+b+c = 29

And if we call "c" the third (and longest) side we can say: c = a+b-1

Now we know one side is 3 longer than another side. But we have to decide is it:

* c = b+3, or

* a = b+3

(a or b can be interchanged, but we have already said c is the side that equals a+b-1)

Now a bit of logic about triangles could probably help us work out which of those two to use. Perhaps if we solved for both possibilities, one of them will be shown to be right and the other wrong.

(Yet to be solved ... anyone?)