As Shan Chen said, the reduction is easy. Note if the DDH adversary gets $g^a,g^b,g^{ab}$ then it gets $g(a,b)$, and if it gets $g^a,g^b,g^{c}$, then he gets a uniformly random string.
Now for the second question. First for a prime number $p$, the order of $\mathbb{Z}^*_p$ is $p-1$. If $G$ is a (integer) cyclic group of prime order $q$, then you are not using $\mathbb{Z}^*_p$, but the order-$q$ subgroup of it ($q$ needs to be large which I assume here).
Why DDH is not hard in $\mathbb{Z}^*_p$? By Euler's criterion, in $\mathbb{Z}^*_p$ there are $\frac{(p - 1)}{2}$ quadratic residues and $\frac{(p − 1)}{2}$ non-residues. So in for a random $x \in \mathbb{Z}^*_p$, the probability of $x$ being a residue (non-residue) is exactly $\frac{1}{2}$. Legendre symbol for an element $x \in \mathbb{Z}^*_p$ can be calculated as:
$$
\Big(\frac{x}{p}\Big)= x^{\frac{p-1}{2}} \bmod p
$$
and the symbol tells whether $a$ is a residue or non-residue:
\begin{equation*}
\Big(\frac{x}{p}\Big) = \begin{cases}
1 &\text{if $x$ is a residue}\\
-1 &\text{if $x$ is a non-residue}
\end{cases}
\end{equation*}
Now given the DDH tuple $(g^a,g^b,g^c) \in (\mathbb{Z}_p^*)^3$, we can calculate the Legendre symbols of $g^a,g^b,g^{c}$. If either $\Big(\frac{g^a}{p}\Big)$ or $\Big(\frac{g^b}{p}\Big)$ is 1, then $Pr[\Big(\frac{g^{ab}}{p}\Big)=1] =1$.
The reason is that a residue $x$ is congruent to a perfect square. So that if $g^a$ is a residue, then it can be expressed as $g^a=h^2$. For $g^{ab}$, it can be rewritten as $(h^b)^2$, which is also a residue.
But for a random $c$ that $c\ne ab$, the probability of it being a residue is only $\frac{1}{2}$, which means we can distinguish $g^{ab}$ and $g^c$ easily.
If the DDH tuple $(g^a,g^b,g^c) \in G^3$, since $q$ divide $p-1$ and $p-1$ is even, $\frac{p-1}{2} = kq$ for some $k$, then $\Big(\frac{g^a}{p}\Big), \Big(\frac{g^b}{p}\Big),\Big(\frac{g^{ab}}{p}\Big), \Big(\frac{g^c}{p}\Big)$ are all 1 (because all elements in $G$ are of order $q$, thus the $q$-th power of all elements in $G$ congruent to 1).
