For the first one you might use the fact that the number of subgroups of a cyclic group of order $n$ equals the number of divisors $d(n)$ of $n$. For a prime $p$, $d(p^k)=k+1$, and hence $k=2$.
For the second statement: pick any $g \in G$, with $g \neq 1$. Then $<g>$ is non-trivial, hence equal to $G$ or a proper subgroup. In the latter case pick $h \in G\backslash<g>$. Then we must have $<h>=G$. So $G$ is cyclic either way.
A subgroup of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ must have order dividing $p^2$ by Lagrange's theorem. Since $p$ is prime, the possible orders of subgroups of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ are $1,p,p^2$. For $1,p^2$ there are only two subgroups of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ with that order, namely $\{(e,e)\}$ and $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ respectively, where $e\in\mathbb{Z}_{p}$ is the identity element.
So now suppose that $A\leq \mathbb{Z}_{p}\times\mathbb{Z}_{p}$ with $|A| = p$. Then since $p$ is prime, $A$ must be cyclic, so there exists some element $(a,b)\in\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ such that $A = \langle(a,b)\rangle $, so $(a,b)$ must have order $p$ in $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$.
The converse is also true, i.e. if $(a,b)$ has order $p$ then the subgroup it generates has order $p$. So the set of subgroups of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ of order $p$ is $\{\langle(a,b)\rangle\mid(a,b) $ has order $p$$\}$.
So now note that the elements of order $p$ of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ are exactly the elements of the form $(a,b)$ where either $a\neq e$, or $b\neq e$ (or both $\neq e$). That is, they are exactly the elements of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}\setminus \{(e,e)\}$, of which there are $p^2-1$. However, each element of order $p$ accounts for $p-2$ other elements of order $p$. You can partition the set of elements of order $p$ into equivalence classes under the equivalence relation $(a,b)\sim(c,d)$ iff there exists $t\in\mathbb{Z}$ such that $(a,b)^{t} = (c,d)$. Each equivalence class has $p-1$ elements (note the identity element of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ is not in the set of elements of order $p$), and so the number of equivalence classes is $\frac{p^2-1}{p-1} = p+1$. So there are $p+1$ subgroups of order $p$.
Now adding to account for the subgroups $\{(e,e)\}$ and $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$, we have that $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ has $p+3$ subgroups.
Best Answer
For $G=Z_{p^2}\times Z_p$, the order $p^2$ subgroups are kernels of nonzero homomorphisms from $G$ to $Z_p$. Such a homomorphism maps $(a,b)$ to $(ta+ub)$ for some $t$, $u\in Z_p$ and so there are $p^2$ homomorphism, including the zero homomorphism But not all non-zero homomorphisms have different kernels. Two have the same kernel iff one is a multiple of the other. So there are $(p^2-1)/(p-1)=p+1$ subgroups of order $p^2$ in $G$.