For part $a)$, use Thomas' hint. You get $ \sum_{i=0}^{n}k(P(X>i-1)-P(X>i)). $
This develops as $P(X>0)-P(X>1)+2P(X>1)-2P(X>2)+3P(X>2)-3P(X>3)+\cdots nP(X>n-1)-nP(X>n)$
for part $b)$:
In general, you have
$\mathbb{E}(X)=\sum\limits_{i=1}^\infty P(X\geq i).$
You can show this as follow: $ \sum\limits_{i=1}^\infty P(X\geq i) = \sum\limits_{i=1}^\infty \sum\limits_{j=i}^\infty P(X = j) $
Switch the order of summation gives \begin{align} \sum\limits_{i=1}^\infty P(X\geq i)&=\sum\limits_{j=1}^\infty \sum\limits_{i=1}^j P(X = j)\\ &=\sum\limits_{j=1}^\infty j\, P(X = j)\\ &=\mathbb{E}(X) \end{align}
$\sum\limits_{i=0}^{\infty}iP(X=i)=\sum\limits_{i=1}^\infty P(X\geq i)=\sum\limits_{i=0}^{\infty} P(X> i)$