The random variable U is uniformly distributed over the interval [3,6]. Find the following probabilities (a) $P[\frac4 5 ≤ U ≤ 4]$ = $$P[\frac4 5 ≤ U ≤ 3]+P[3≤ U ≤ 4]=0+P[U ≤ 4]=\frac{(4-3)}{(6-3)}=\frac1 3$$ Is this even right? I'm just not so confident.
(b) $P[U > 5]$ = $$1-P[U ≤5]=1-\frac{8-3}{6-3}=1-\frac5 3$$
The answer for b is a positive number according to my friends so I am not sure that I'm correct in this one either.
(c) $P[16 ≤ U^2 ≤ 36]$= 
First I simplify them
$$P[4≤ U ≤ 6]  or  P[-6 ≤ U ≤ -4]$$
Then I form the equation
$$P[16 ≤ U^2 ≤ 36]= P[4≤ U ≤ 6]+P[-6 ≤ U ≤ -4]$$
But I am not sure what to do afterwards, do I have to integrate it? If so, how??
(d) $P[4−2|U|≥−8$= $$ P[|U|≤ 5] = P[−5 ≤ U ≤ 5] = P[3 ≤ U ≤ 5] = 1/3$$ Is this even correct??
Trying to do the exercises based on my friend's notes on the classes. So far I still feel lost on these kinds of exercises but the answers just seem so off so far so I would appreciate some advice.

Note that your probability density function is  $f(x)= 1/3$ for $x\in[3,6]$ and $f(x)=0$, otherwise.
$(a)$ $$P[\frac4 5 ≤ U ≤ 4]= P[ 3 ≤ U ≤ 4]=1/3$$
$(b)$ $$P[U > 5]=P[5<U<6]=1/3$$
$(c)$ $$P[16 ≤ U^2 ≤ 36]= \\P[4≤ U ≤ 6]+P[-6 ≤ U ≤ -4]=\\2/3+0=2/3$$
