Do all periodic waves have a fundamental frequency? Do all periodic wave forms have a fundamental frequency (if true, why?) and I was also wondering if a pulse also has a fundamental frequency? 
 A: Let's first look at a definition of "fundamental frequency": "The fundamental frequency, often referred to simply as the fundamental, is defined as the lowest frequency of a periodic waveform." Given this definition, all periodic wave forms definitely have a fundamental frequency $f$, which is exactly the reciprocal of the period $T$ of the periodic waveform $$f=\frac {1}{T} \tag 1$$ @ThePhoton has pointed this out correctly in his above comments. The reason is that any periodic waveform $$f(t)=f(t+T)\tag 2$$ with period $T$ can be expanded in a Fourier series with period $T$: $$ f(t)=\sum_{n=-\infty}^{+\infty} c_n \exp( i\frac {2\pi nt}{T}) \tag3$$ where the coefficients are given by $$c_n=\frac {1}{T}\int_{t_0}^{t_0+T} f(t)\exp(-i \frac {2\pi nt}{T}) \tag 4$$ and the integral is over any period of the function starting at $t_0$. Thus the fundamental frequency waveform $f_1 (t)$ of an arbitrary periodic waveform $f(t)$ is given by the Fourier term for $n=1$: $$f_1(t)=c_1 \exp( i\frac {2\pi t}{T}) \tag 5$$ with $c_1=c_1*$ defined by eq. (4). A single pulse does't have a fundamental frequency in this sense. It can, however, be decomposed into infinitely many frequency components that correspond to the Fourier transform of the single pulse. 
A: It depends on what you mean. 
If you're feeling mathy, you might see periodic waves and think periodic function, $f(x) = f(x + p)$, and fundamental frequency and think 'if it doesn't have a fundamental, it's not periodic.' In math, given $f(x) = f(x+p_1)$, and $g(x) = g(x+p_2)$, $h(x) = f(x) + g(x)$ is not periodic unless $\frac{p_1}{p_2} \in $ rationals$^1$, despite $h(x)$ being composed of periodic functions. QED. Yes.
If you were coming from a musical or electronics background, you might see 'periodic wave' as a 'signal' with periodic properties - this makes more sense given the OP asks about fundamental frequencies which are a common topic for musicians. Here, the fundamental is the tone which the human ear identifies as the specific pitch of the signal (because it's the loudest [lowest] psycho-acoustically).
The one dimensional wave equation, with its famous set of 'overtones', approximates string, brass and wind instruments. The fundamental is the lowest frequency in the harmonic series. Round percussion and square percussion have different, more complicated overtones. For this reason and decay, describing these signals by their fundamentals may be less useful, but they do have a lowest frequency which usually decays more slowly than the higher frequency components. Not necessarily.
Incidentally, we tune most pianos to equal temperament which means that adjacent notes have a ratio of $\root{12}\of{2}$. To generate a signal without a fundamental frequency, simply play any two notes that aren't integer octaves apart. Trivially, testably no.
If you had a background in wavelets, solitons or acoustics, you might include periodic travelling waves$^2$. OP, is this what you mean when you say 'pulse'? Here, the only requirement is that the frequency components have similar speeds in the medium. These probably aren't periodic functions because a wavelet needn't repeat. A pulse will have a 'lowest frequency', but the pulse's duration may be too short to identify that frequency - see ears and NSST. Still no.
Last, if we're talking about sound and being pedantic, our ears have hairs that vibrate at different frequencies, so rather than 'no sound is periodic', every sound is necessarily periodic. If you managed to play Dirac's delta function, first of all, wow and nice explosion, but your ear would hear it's own frequency response characteristic. This and hitting a system with a Heaviside Step function are techniques for quickly examining a system's frequency response.
$^{^1\text{Irrationals are uncountable while rationals are merely infinite. Given 2 random frequencies, the signal's assuredly non-periodic.}}$
$^{^1\text{Travelling waves exist with open and closed boundary conditions.}}$
