Question 633336
Good job with the definition of the function. The <= , and the (1/2)x make me pretty sure that I understand what you mean.
As I understand, the function is defined in pieces:
{{{f(x)=3}}} for {{{x<=-2}}} and
{{{f(x)=(1/2)x+6}}} for {{{x>2}}}
and you need to calculate the value of {{{f(x)}}}
for {{{x=-5}}}, {{{x=-2}}}, {{{x=0}}} and {{{x=2}}}.
 
{{{x=-5}}}, and {{{x=-2}}} are in the first part of the domain,
the part where {{{x<=-2}}}.
In that part, {{{f(x)=3}}}, so
{{{highlight(f(-5)=3)}}} and {{{highlight(f(-2)=3)}}}
 
{{{x=0}}}, and {{{x=2}}} are in the other part of the domain,
the part where {{{x>-2}}}.
In that part, {{{f(x)=(1/2)x+6}}}, so we enter the x value into {{{f(x)=(1/2)x+6}}} to calculate the {{{f(x)}}} values:
For {{{x=0}}}, {{{f(0)=(1/2)*0+6=0+6=6}}} so {{{highlight(f(0)=6)}}} 
For  {{{x=2}}}, {{{f(2)=(1/2)*2+6=1+6=7}}} so {{{highlight(f(2)=7)}}}
 
The problem does not ask for a graph, but I like to draw these pictures.
The graph of that piecewise function looks like this:
{{{drawing(300,300,-6.5,3.5,-2,8,
grid(1),
blue(circle(-2,3,0.04)),blue(circle(-2,3,0.08)),
blue(circle(-2,3,0.12)),
blue(arrow(-2.08,3,-6,3)),blue(arrow(-2.08,3.03,-6,3.03)),
blue(arrow(-2.08,2.97,-6,2.97)),
blue(arrow(-1.8,5.1,3.6,7.8)),blue(arrow(-1.8,5.12,3.6,7.82)),
blue(arrow(-1.8,5.08,3.6,7.78)),
circle(-2,5,0.2),
green(circle(-2,3,0.3)),green(circle(-5,3,0.3)),
green(circle(-0,6,0.3)),green(circle(2,7,0.3))
)}}} I drew the function in blue and put green circles around the 4 points (x,f(x)) that we calculated.