A periodic function repeats over and over.  Since we are not given any values
for f(x), we can only hope that f(2), f(-4), and f(7) all turn out to be repeats
of the same value.

Since f(x) has period 0.3, 

f(x) = f(x+0.3n) for any integer n.

Therefore,

f(-4) = f(-4+0.3n)

We wonder if there is an integer n so that f(-4) = f(2),
for then we could find their common value, so to find out,
we set -4+0.3n = 2
          0.3n = 6
             n = 6/0.3     
             n = 20

YES!! There is such an integer, n=20, so f(2) = f(-4+20∙0.3) = f(-4)

f(2)= 5f(-4)+2
f(2)= 5f(2)+2
-4f(2)=2
f(2)=-0.5

We want to find f(7),  So, as before

f(7) = f(7+0.3n) for any integer n

As before, we wonder if there is an integer n so that f(7) = f(2),
for then we could find their common value, so to find out,
we set 7+0.3n = 2
         0.3n = -5
            n = -5/0.3     
            n = -50/3

Oh shucks! That's not an integer, so it looks at first like we're stuck!
But wait! All may not be lost!  Think now! What was given that we haven't used?
Aha! Eureka! To the rescue, we were given that the function is even, and that
means

f(-x) = f(x) for all x.

Therefore f(2) = f(-2)

So it will be just as good if there is an integer n so that f(7) = f(-2),
for then we could find their common value, so to find out if there is such an
integer, we set 7+0.3n = -2
                  0.3n = -9
                     n = -9/0.3     
                     n = -30

Hurray! -30 is an integer. So that means 

f(7) = f(7-30∙0.3) = f(-2) = f(2) = -0.5   <--the answer

Edwin