Question 111021
I cannot answer your question about a special word problem section. That is the Webmaster's area.

As a volunteer tutor, I can say that I am not inclined to take on answering a question that has 3 questions embedded in it. I might have time to answer 1 or 2, but perhaps not all 3. Also, by posting 3 questions, you could get lucky and have 3 tutors work in parallel to give you the 3 answers in 1/3 the time.

As to setting up the problems:

1) The consecutive integers would be x, x+1, and x+2. And we are told that 
x + (x+1) is 11 more than x+2. So,
2x+1=x+2+11
x=12.

2) A rectangule has area A=w*l, where w=width, l=length. 
We are told length is 15m more than w, so we could say l = w+15, or l-15=w.
That means the original area A = w(w+15). We don't know what A is though.
But the 0.5m row of bushes, means the length is increased by 1m and the width is increased by 1m (.5 is added on all sides). When that happens, A is increased by 56.
Letting B = new area, the equation is
B = (w+1)(w+15+1)
A = w(w+15)
B = A +56
So...
(w+1)(w+16) = w(w+15)+56.

3) With money problems, you have to keep track of the count as well as the value of the coins involved. Note that we can call the number of dimes d and the value, in cents, will be 10d. Likewise, nickels can be called n and their value is 5n. Pennies are p and have value p (or 1p, but the 1 is superfluous).

The total number of coins will be d+n+p. Their value will be 10d+5n+p.
We know 10d+5n+p=50 cents. We also know the number of dimes is 3, so we can say
30+5n+p=50. Subtracting 30 from both sides, we have 5n+p=20.

We are told she has half as many nickels as pennies. That means, if we have 2 pennies, we have 1 nickel; 4 pennies, 2 nickels; 6 pennies, 3 nickels; 8 pennies, 4 nickels; etc.

In all cases, the relationship is p=2n.

That means
5n+2n=20

7n = 20, which presents us with a problem. 7 does not go into 20 an even number of times, so there is no solution.