I'm not sure how to find the expected value.
If I pick 5 cards out of 52 there are 52 choose 5 ways of drawing 5 cards which is 2,598,960 ways.
There are 13/52 spades in deck.
I'm not sure whether I'm along the right track. Any help would be appreciated. Thank you in advance.
Use the additivity of expectation: expectation of sum of RVs is the sum of their expectations.
Let $X_i,~i=1,\dots,5$ be RVs with $X_i = 1$ if the $i$-th card is a spade and $=0$ otherwise. Then these RVs are identically distributed (but not independent), and each is Bernoulli with parameter $1/4$ (fourth of the cards are spades), so $E[X_i] = 1/4$.
Why the same distribution? The number of ways to select $5$ cards with first being a spade is exactly the same number of ways to select $5$ cards with, second a spade (any choice of the former type becomes a choice of the latter type by swapping the first and second cards), which is the same as number of ways to select $5$ cards with third being a spade, etc.
The number of spades $N = X_1+\dots + X_5$, and
$$ E[N] = E [ X_1 + \dots + X_5] = E[X_1] +\dots + E[X_5] = 5 \times \frac{1}{4}=\frac54.$$
Hint: For $k = 1,2,3,4,5$, let $X_k$ be $1$ if the $k$-th card you drew is a spade, and $0$ otherwise. Then, $$X = X_1+X_2+X_3+X_4+X_5$$ and then by using linearity of expectation, $$E[X] = E[X_1+X_2+X_3+X_4+X_5] = E[X_1]+E[X_2]+E[X_3]+E[X_4]+E[X_5].$$
Also, for each $k$, we have $$E[X_k] = 1 \cdot P[X_k = 1] + 0 \cdot P[X_k = 0] = P[\text{the} \ k\text{-th card is a spade}].$$
Can you take it from here? This should be easier than computing the probabilities of drawing exactly $0,1,2,3,4,5$ spades.
You can calculate the individual probabilities: $$P(\text{n spades in 5 draws})=\frac{{39 \choose 5-n}{13 \choose n}}{{52 \choose 5}}$$
So the expected value would be
$$E(\text{# of spades})=\sum_{n=0}^{5}n.P(n)=\sum_{n=0}^{5}n.\frac{{39 \choose 5-n}{13 \choose n}}{{52 \choose 5}}$$
