Question 1007076
<pre>
I'll just do the first two, as that's all the time I have. But
hopefully you'll catch on to how to do it yourself.

There is a table at this site:

https://www.easycalculation.com/statistics/normal-ztable.php 

On that site, for z = 1.00, we find 1 in the far left column and read
0.3413 just next to the right of it.  It is in the column headed 0.
This table reads right from the middle of the normal curve. What we 
are reading off the table is the area 0.3413 on the graph below, 
which is the area between the y-axis and the green line at z=1.  But 
we want the proportion of area to the right of the green line, so 
we subtract 0.3413 from one-half or 0.5000 and get 

 0.5000
-0.3413
-------
 0.1587  <-- the proportion of area to the right of the green line
             below at z = 1

{{{drawing(500,250,-5,5,-.5,1.5, graph(500,250,-5,5,-.5,1.5, exp(-x^2/2)), 
locate(.01,.5,0.3413), locate(1.01,.15,0.1587), locate(-1.1,.4,0.5000),
green(line(1,0,1,exp(-1^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z) 
)}}}

Answer: 0.1587

--------------------

Next we want the proportion of area to the right of -1.05.  On the table
we look up 1 on the far left column and then go over to the column headed
0.05, and read 0.3531.  That reads the area between the green line below
marked at z=-1.05 and the y-axis.  But we also want the whole right half
of the area so we must add 0.5000 to that 0.3531

 0.3531
+0.5000
-------
 0.8531


{{{drawing(500,250,-5,5,-.5,1.5, graph(500,250,-5,5,-.5,1.5, exp(-x^2/2)), 
 
green(line(-1.05,0,-1.05,exp(-1.05^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z),
locate(-1, .5,0.3531),locate(.2,.4,0.5000)
  
)}}} 

So the total proportion of area to the right of the green line at
z=-1.05 is 0.8531

Notice that for the c part, you don't need to look in the table, because
half the proportion of area is to the right of z=0, so the answer is
0.5000.  No need for a table for that one.

If you have any more questions, ask me in the thank-you note form below.
I'll get back to you by email.  BTW I don't charge any money. I do this
for fun! 

Edwin</pre>