Show the following including $\sigma$ function 
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*How do I show that $\sigma (2k)=4k$ if and only if $k=2^{p-2}(2^p-1)$ where $2^p-1$ is a prime number.

*I want to show that if $k$ is odd and $\sigma(k) = 2k$ then $k=p^am^2$ for some p with $(p,m)=1$ and $p=a=1$(mod 4).
I know that $\sigma (2k)=\sum \limits_{d|2k} d$, where I can see that d|2...

for (1) (it should help you with ideas for (2) as well)
$\sigma(n)$ is the sum of divisors of $n$. In your case $n=2k=2^{p-1}q$ where $q=2^p-1$ itself being a prime. So all possible divisors of $n$ will be of the form $2^aq^b$, where $a \in \{0,1,2, \ldots , p-1\}$ and $b \in \{0,1\}$. So
\begin{align*}
\sigma(2k) & =\sum_{a=0}^{p-1}2^a+\sum_{a=0}^{p-1}2^aq\\
& = \frac{2^p-1}{2-1}+q\left(\frac{2^p-1}{2-1}\right)\\
&=(2^p-1)(q+1)\\
&=2^p(2^p-1))\\
& =2^22^{p-2}(2^p-1)\\
& = 4k
\end{align*}
