A fuel with a composition of C=86%, H=14%, is burned in dry air. The
analysis of the combustion products indicate on a dry basis(after
condensing and removing all water), O2=1.5%, CO=600ppm.
Determine the equivelance ratio of combustion?

This sounds very much like a homework problem and google answerers and
commenters are discouraged from answering such questions.  Instead of
giving you an answer, I will walk you through what you need to do.

First thing is to determine what hydrocarbon we're talking about. 
Hydrocarbons can only fit certain carbon-hydrogen ratios. 
Specifically, the ratio is H = 2C +2.  So, for instance, propane has 3
Carbons.  2 x 3 + 2 = 8, meaning it must have 8 hydrogens.  And so
forth.  Recalling that carbon has an atomic mass of 12 while hydrogen
has an atomic mass of 1 will allow you to set up a simple system of
equations to determine what hydrocarbon we're taling about.

Check to make sure the 86% v 14% is realistic.  I did my own
calculations and end up with a rather hefty hydrocarbon; one that is
not likely to be found in nature.

Anyways, once you figure out the hydrocarbon, you need to do the
stoichiometry.  Figure out how much oxygen is required to burn said
fuel.  Add CO and remove the water from the products side and then
it's just some simple math.

Good luck!