Constructing pullback and pushout problem

i) Let $p$ be a prime and $f: \Bbb Z \rightarrow \Bbb Z_p$ and $g: \Bbb Z_{p^2}\rightarrow\Bbb Z_p$ be the canonical epimorphism. Show that the pullback of $f$ and $g$ is isomorphic to $\Bbb Z \oplus \Bbb Z_p$

ii) $f: \Bbb Z \rightarrow \Bbb Q$ the embebbing and $g$: $\Bbb Z$ $\rightarrow$ $\Bbb Z$ the multiplication for $n>1$. Show that the pushout of $f$ and $g$ is isomorphic to $\Bbb Q \oplus \Bbb Z_n$

My idea of to prove this is to show that since the natural pullback of f and g is $\Bbb Z_{p^2} \times \Bbb Z$ so by showing $\Bbb Z \oplus \Bbb Z_p$ is also pullback of f and g then both are isomorphic since the pullback is unique (the same idea to prove ii)

$\Bbb Q$ $\oplus$ $\Bbb Z_n$

Your idea is not quite right: the product $M\times N$ of two modules $M,N$ is the pullback of $M\to 0\leftarrow N$, but in your problem you are interested in the pullback of ${\mathbb Z}_{p^2}\to {\mathbb Z}_p\leftarrow {\mathbb Z}$, which can be realized as the submodule of ${\mathbb Z}_{p^2}\times{\mathbb Z}$ consisting of those pairs $(\overline{x},y)$ such that $x\equiv y$ modulo $p$.
Given such a pair, consider the pair $(\overline{x-y},y)$; since $x\equiv y$ modulo $p$, the class $\overline{x-y}$ then belongs to the kernel of ${\mathbb Z}_{p^2}\to {\mathbb Z}_p$ - do you know how the latter looks like? Conversely, given a pair $(\overline{x^{\prime}},y^{\prime})$ with $\overline{x^{\prime}}\in\text{ker}({\mathbb Z}_{p^2}\to {\mathbb Z}_p)$ and $y^{\prime}\in {\mathbb Z}$, how could you reverse the above transformation to get an element $(\overline{x},y)\in {\mathbb Z}_{p^2}\times {\mathbb Z}$ with $x\equiv y$ modulo $p$?
Said differently, you can note that the desired pullback, call it ${\mathbb Z}_{p^2}\times_{{\mathbb Z}_p} {\mathbb Z}$, fits into a short exact sequence $$0\to \text{ker}({\mathbb Z}_{p^2}\to{\mathbb Z}_p)\to {\mathbb Z}_{p^2}\times_{{\mathbb Z}_p} {\mathbb Z}\to {\mathbb Z}\to 0,$$ and any short exact sequence ending in ${\mathbb Z}$ splits.