Let x = the number of years ago it was when Harry was a year 
younger than Tom is now.

Then

h-x = t-1 

Dick was d-x then.

And Tom is twice that age now, so

t = 2(d-x)

t = 2d-2x

Dick is half as old as Harry will be when Tom is ten years 
older than Dick is now.

Let y be the number of years when Tom is ten years older 
than Dick is now.

Then 

t+y = d+10

Then in y years Harry will be h+y and Dick's age is half 
of that, so
 
d = (h+y)/2 or clear of fractions:

2d = h+y

Harry is one and a half times as old as Tom was when Dick 
was ten years younger than Harry is now. 

Let z = the number of years since Dick was ten years 
younger than Harry is now.

Then

d-z = h-10

Tom was t-z

Harry is 1.5 times that.  So

h = 1.5(t-z)

or multiplying both sides by 2 to avoid fractions

2h = 3(t-z)

And distributing to avoid parentheses:

2h = 3t-3z

We have this system of 6 equations in 6 unknowns

h-x = t-1
t = 2d-2x  
t+y = d+10
2d = h+y
d-z = h-10
2h = 3t-3z

Solve that either by subtitution or elimination
and get

t=18, d=16, h=24, x=7, y=8, z=2 

Checking:

Tom is twice as old as Dick was when Harry was one year 
younger than Tom is now.

Tom is 18. 7 years ago Harry was 17, which is 1 year 
younger than Tom is now. And 7 years ago Dick was 9, 
and Tom is indeed twice that. 

Dick is half as old as Harry will be when Tom is ten 
years older than Dick is now.

In 8 years, Tom will be 26, and will indeed be 10 years 
older than Dick is now. 

In 8 years, Harry will be 32, and indeed Dick is half 
that now.

Harry is one and a half times as old as Tom was when 
Dick was ten years younger than Harry is now.

2 years ago Dick was 14 and that is 10 years younger 
than Harry's present age of 24.

2 years ago Tom was 16 and Harry is 1.5 times that or 24.

So everything checks.

Edwin