Question

Both questions please

Answer 1

Let there be $h$ 50 cent coins (call them halfs), $d$ 10 cent coins (dimes) and $n$ 5 cent coins (nickles).  

81 coins:

$h+d+n=81$

Three more dimes than halfs:

$d=h+3$

Nickels are twice halfs plus dimes:

$n=2(h+d)$

Three equations in three unknowns.  We'll eliminate $n$ first:

$n=81-(h+d)=2(h+d)$

$81=3(h+d)$

$h+d=27$

Substituting $d=h+3$

$h+h+3=27$

$2h=24$

$h=12$

$d=h+3=15$

$n=2(h+d)=2(27)=54$

Check we have the right number of coins:

$h+d+n=12+15+54=81 \quad\checkmark$

Good.   The amount of money is

$50h+10d+5n=0.50(12)+0.10(15)+0.05(54)=\ 10.20$

Part 2.

a for amount Tim gave cousin.  Initially Tim has $146.65, cousin $10.55

After he gave the money Tim had twice as much:

$146.65 - a = 2(10.55 + a)$

$146.65 - a = 21.10 + 2a$

$146.65 - 21.10 = 3a$

$a = \dfrac{146.65 - 21.10}{3}= 41.85$

That's the answer to b.  After the transfer the cousin has $10.55+41.85=\ 52.40$