1. Distribution of $N$

Since the coin has probability of head $P(H)=\frac14,$ and it is tossed until the first head appears, $N$ follows a geometric distribution: $P(N=n)=\left(\frac34\right)^{n-1}\left(\frac14\right), \qquad n=1,2,3,\dots$

2. Condition for no real roots

Given the quadratic equation $64x^2+5Nx+1=0,$ it has no real root when its discriminant is negative: $D=(5N)^2-4\cdot 64\cdot 1<0.$ So, $25N^2-256<0$ $25N^2<256$ $N^2<\frac{256}{25}.$ Thus, $N<\frac{16}{5}=3.2.$ Since $N$ is a positive integer, this means $N=1,2,3.$

3. Required probability

So we need $P(N\le 3)=P(N=1)+P(N=2)+P(N=3).$

Using the geometric distribution: $P(N=1)=\frac14,$ $P(N=2)=\frac34\cdot \frac14=\frac{3}{16},$ $P(N=3)=\left(\frac34\right)^2\cdot \frac14=\frac{9}{64}.$

Hence, $P(N\le 3)=\frac14+\frac{3}{16}+\frac{9}{64}.$ Taking LCM $64$: $P(N\le 3)=\frac{16+12+9}{64}=\frac{37}{64}.$

Thus, $\frac{p}{q}=\frac{37}{64}$ with coprime $p=37$, $q=64$.

4. Find $q-p$

$q-p=64-37=27.$

Final Answer

$\boxed{27}$

The derived answer matches the stored correct answer.