2^x+2^x+2^x+2^x= 4*2^x=2^2*2^x=2^(2+x)
thus 2^(2+x)=2^7
2+x=7
x=5
What you have provided are not solutions, but answers. Solutions have all the working involved to get to the answer.
Anyway, for Q1
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This equation is actually not solvable using traditional methods, however, you can tell that since the LHS needs to be a multiple of 2, that means must be odd. So you can just substitute odd integers until you get LHS = RHS.
In this case, works.
There are many equations that pop up which don't have a method to solve them other than numerical methods or guessing. This is one. In fact, there are many equations that don't have an exact solution at all and only decimal approximations are possible. I suspect you were given this question not to practice solving methods, but just to get you thinking.