Yes your approach is indeed correct and is probably the right way to go.
For the first problem, all we need is that $mx + ny = r$ where $m,n,r \in \mathbb{Z}$ has integer solutions iff $(m,n) | r$. Since $(a,b) = 1$, there exists $x,y \in \mathbb{Z}$ such that $ax + by = 1$. Since $c|a$, we have $a = ck_1$ and since $d|b$, we have $b = dk_2$. Hence, we get that $ck_1x + dk_2y = 1.$Hence, $(c,d) = 1$.
For the second problem,
As before, since $(a,b) = 1$, we have $x_1,y_1,x_2,y_2 \in \mathbb{Z}$, such that $ax_1 + by_1 = 1$ and $ax_2 + cy_2 = 1$. Multiplying both out we get that $(ax_1 + by_1)(ax_2 + cy_2) = 1$ i.e. $a(ax_1x_2 + by_1x_2 + cy_2x_1) + bc (y_1y_2) = 1$. Note that $(ax_1x_2 + by_1x_2 + cy_2x_1), y_1y_2 \in \mathbb{Z}$. Hence, we have $(a,bc) = 1$.
For the third problem,
We will first prove that $(a^n,b) = 1$ for all $n \geq 1$. This is proved by induction. Clearly, for $n=1$ it is true since we are already given that $(a,b) = 1$. Assume that it is true for all $n$ up to $m$ i.e. we have $(a^n,b) = 1$, $\forall n \in \{1,2,\ldots,m \}$. Hence, we have that $ax + by = 1$ and $a^m x_1 + b y_1 = 1$ for some $x,x_1,y,y_1 \in \mathbb{Z}$. Multiplying the two out, we get that $(ax + by) \times (a^m x_1 + b y_1) = 1$ i.e. $a^{m+1} (xx_1) + b (axy_1 + a^n y x_1 + byy_1) = 1$. Hence, we get that $(a^{m+1},b) = 1$. Hence, by induction, we have that $(a^n,b) = 1$, $\forall n \in \mathbb{N}$. Now fix an $n \in \mathbb{N}$, we need to prove that $(a^n,b^k) = 1$, $\forall k \in \mathbb{Z}^+$. This is done by inducting on $k$ as before i.e. call $a^n = A$ and then induct as before to prove that $(A,b^k) = 1$ $\forall k \in \mathbb{N}$, given that $(A,b) = 1$. Hence, we get that $(a^n,b^k) = 1$, $\forall k,n \in \mathbb{N}$, given $(a,b) = 1$.