(a) The set of natural numbers with the operation of subtraction does not form a group because the following properties of a group are not present:
Closure: If $a$ and $b$ are natural numbers and $a<b$, then $(a-b)$ is not a natural number.
Associativity: $(a-b)-c\ne a-(b-c)$, e.g. $(20-12)-5\ne 20-(12-5)$.
Identity: 0 is an integer, and $a+0=0+a=a$. True (So the identity $e=0$)
Inverses: $a+(-a)=(-a)+a=0$. If $a$ is a natural number, $(-a)$ is not a natural number.
(Andrei)

(b) Rational numbers with multiplication A rational number, by definition is expressible as $m/n$ where $m,n$ are integers and $n\ne 0$. With the operation of multiplication, the following occurs: $m/n*t/y=\mathrm{mt}/\mathrm{ny}$. Once again, both the numerator and denominator are integers (also $n,y\ne 0$), hence closure is achieved.

When one investigates division, the following may occur. Let t=0 (it is a numerator so this integer is allowed)

${\displaystyle (m/n)/(t/y)=m/n*y/t=(\mathrm{my})/(\mathrm{nt}).}$

But now the denominator is equal to 0. Any real number divided by 0 is not defined within the set of rational numbers. Therefore, condition (1) is contravened; closure is not achieved.

Also division is not associative e.g. $(100/50)/5\ne 100/(50/5).$
(Curt)

(c) Positive integers with multiplication I check all the properties of the groups for the set of positive integers with the operation of multiplication:
Closure: If $a$ and $b$ are positive integers, then $a*b$ is also a positive integer. True.
Associativity: $(a*b)*c=a*(b*c)$ for all $a,b,c$ in the set. True.
Identity: 1 is a positive integer, and $a*1=1*a=a$. True (So the identity e = 1).
Inverses: $a*(1/a)=(1/a)*a=1$. But $1/a$ is not always an integer.

So, the set of positive integers with the operation of multiplication does not form a group as it does not contain inverse elements.
(Andrei)

(d) The set of all positive even integers with the operation of multiplication would not form a group because this set does not contain inverses. But it is noted that no identity exists in this set. For any real number, the identity element of multiplication is 1. But 1 is not an even integer, and therefore not a member of the set of all even integers. This contravenes condition (3).
(Curt)

(e) The set of integers with the operation * defined such that $m*n=m+n+1$ is a group. The identity $e$, is defined as being an element such that $m*e=m$ or (as all the operations that make up * are not affected by order) $e*m=m$. Therefore $m*e=m=m+e+1$ which implies $e+1=0$ and so $e=-1$.

The inverse element, $m^{-1}$, has the property that $m*m^{-1}=e=-1=m+m^{-1}+1$ an so $m^{-1}$, the inverse of $m$, is given by $m^{-1}=-(m+2).$
(Curt

(f) The set of integers with the operation * defined by $m*n=m+(-1{)}^{m}n$ is a group with the following properties:
Closure: If $m$ and $n$ are integers, then $m*n$ is also an integer. True: $m*n=m+n$ when $m$ is even and $m-n$ when $m$ is odd.
Associativity: $(m*n)*p=m*(n*p)$. True.
Identity: $m*e=m+(-1{)}^{m}e=m$ for all integer $m$ implies $e=0$ and also $0*m=0+(-1{)}^{0}m=m$, so the identity element is 0.
Inverses. One must work separately for even and odd elements. If $m$ is even ( $m=2p$ where $p$ is an integer) then $m*m^{-1}=0$ implies $m+m^{-1}=0$ and so $m^{-1}=-m.$ Similarly $-m*m=0$ which verifies that the inverse is $-m$. If $m$ is odd then the inverse of $m$ is $m$.
The inverse of $m$ is given for both cases by $m^{-1}=(-1{)}^{m+1}m$.
(Andrei)

(g)The set of all real numbers excluding only the number -1 together with the operation $x*y=\mathrm{xy}+x+y$ is a group.

The inverse $e$ is such that $x*e=x=\mathrm{xe}+x+e$ and so $\mathrm{xe}=-e$. Therefore $e=0$ is the identity.

If $x*x^{-1}=0={\mathrm{xx}}^{-1}+x+x^{-1}=0$ then $x^{-1}(x+1)=-x$ and so the inverse of $x$ is given by $x^{-1}=-x/(x+1)$ (hence $x\ne -1$, or else $x^{-1}$ is not a real number, which would contravene the properties of the group)
(Curt)