$G$ is a group dan $K$ is a normal subgroup of $G$. If $H$ is a subgroup of $G$ then there is a one on one correspondence between every elemen in $H/K$ and ${H}/{H\cap K}$, so that $HK/K\approx H/H\cap K$.
With Fundamental Theorem of Homomorphism, I want to prove that
a. $K$ is a normal subgroup of $HK$
b. $H\cap K$ is a normal subgroup of H
c. There is $\psi : H \to HK/K$ with $\psi(a)=Ka$, $a \in H$ is a homomorphism from group $H$ onto $HK/K$
d. $Ker(\psi)=H\cap K$
For a. I proved them by proving $HK$ is normal in $G$ first then proving that since $K$ is normal in $G$ and $K \subseteq HK \subseteq G$, we have that $K$ is normal in $HK$.
For b. Since $H$ and $K$ are normal in $G$, given any $g\in G$, $gHg^{-1} \subseteq H$ and $gKg^{-1} \subseteq K$. So we have $g(H\cap K)g^{-1} = gHg^{-1} \cap gKg^{-1} \subseteq (H\cap K)$. So $(H\cap K)$ is normal in $G$. Since $(H\cap K) \subseteq H \subseteq G$, we have that $(H\cap K)$ is normal in $H$.
I have no idea for how to prove c and d.
Well, every element of $HK/K$ is a coset of the form $(hk)K$ $(=K(hk)$ since $K$ is normal in $HK)$ where $h \in H$ and $k \in K$. So, the function $\psi : H \to HK/K$ such that $$(\forall a \in H) \quad \psi(a) = aK \,(=(a1)K \in HK/K)$$ at least makes sense. Now, if $h \in H$ and $k \in K$, note that $hkK = hK$ because $h^{-1}(hk) = k \in K$. Thus, every element of $HK/K$ is in fact of the form $hK = \psi(h)$ for some $h \in H$, that is, $\psi$ is surjective. Now, $(ab)K = (aK)(bK)$ for every $a,b \in H$ tells us that $\psi$ is a group homomorphism, and, since $K$ is the identity element of $HK/K$, we also have that $$\ker \psi = \{a \in H : aK = K\} = \{a \in H : a \in K\} = H \cap K.$$