Your answer ($0.35$) looks correct, and the textbook answer is wrong.
The fraction $\frac{p(0)+p(1)+p(2)+p(3)}{4}$ will evaluate to $\frac{1}{4}=0.25$ for any probability mass function $p$, so that particular ratio does not have any significance for the expectation of $X$.
A probability density function $f$ must satisfy:
1) $f(x)\ge 0 $ for all $x$,
and
2) $\int_{-\infty}^\infty f(x)\, dx =1$.
Your density has the form $$f(x)=\begin{cases}c \cdot x^{-a} & x\ge x_l \\ 0 & \text{ otherwise}\end{cases}$$
where $x_l>0$.
We need 1) to hold; $f$ must be non-negative.
When does that happen?
The first thing to note here is that, since $x_l>0$, it follows that $x^{-a}\ge0$; and thus $c$ must be positive in order for 1) to hold.
So far so good. $a$ can be any number (so far as we have surmised) and, for $c>0$, $f$ would define a density as long as condition 2) holds.
Your task now is to figure out when it does.
A hint towards achieving that end would be to consider when the integral appearing in 2) is converges. If the integral does converge, you can then select $c$ so that it converges to 1; and in this case, $f$ would indeed define a density.
If the integral does not converge, then $f$ would not define a density.
Read no further if all you want is a hint...
To determine the range of values of $a$ for which $f$ is a density we need to determine when $$\tag{3}\int_{x_l}^\infty c x^{-a}\,dx$$ converges.
Towards this end, note that the integral in (3) is convergent if and only if $a>1$. This is because the $p$-integral $\int_{x_l}^\infty {1\over x^p}\,dx $ converges if and only if $p>1$ (the lower limit presents no problems, since $x_l>0$).
This answers your question as to what range of values of $a$ (I assume $a$) give a valid density.
If you have $a>1$ and want to find the value of $c$, use 2): set
$$
1=\int_{x_l}^\infty cx^{-a}\,dx
=\lim_{b\rightarrow\infty} { -cx^{-a+1}\over -a+1}\biggl|_{x_l}^b={cx_l^{1-a}\over a-1},
$$
then solve for $c$.
Best Answer
For part (iii).
From this post, you have $$ f_X(x) =2^{a-1}(a-1)x^{-a}, \ x\ge 1/2. $$ Then $$\eqalign{ p(k)&= \int_{k-{1\over2}}^{k+{1\over2}} 2^{a-1}(a-1)x^{-a}\,dx\cr &\approx \Bigl( (k+{1\over2}) - (k-{1\over2}) \Bigr) 2^{a-1}(a-1)k^{-a}\cr & = 2^{a-1}(a-1)k^{-a}.} $$
I'm not sure if this is what you're expected to do, as the integral above can be computed exactly (and this is what you do for part (i)).
Perhaps you're expected to think of the pmf as a "bar chart". The bar chart can be approximated by the graph of the density function. Each bar has width 1, is centered at $k$, and the height is $f_X(k)= 2^{a-1}(a-1)k^{-a}$.
(I'm not sure what "for small $a$" means here, $a$ must be greater than 1 in order for $f$ to actually define a density.)