Disregard my earlier question.
Recall that the variance of a random variable [tex]X[/tex], denoted [tex]\mathbb V(X)[/tex], is given by
[tex]\mathbb E((X-\mu)^2)[/tex]
where [tex]\mathbb E(X)[/tex] denotes the expected value/mean of [tex]X[/tex], and [tex]\mu=\mathbb E(X)[/tex] is the actual mean of [tex]X[/tex].
Now, recall that
[tex]\mathbb V(X)=\mathbb E((X-\mu)^2)=\mathbb E(X^2-2\mu X+\mu^2)[/tex]
[tex]\mathbb V(X)=\mathbb E(X^2)-2\mu^2+\mu^2[/tex]
[tex]\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2[/tex]
(a) [tex]Z=2+10X[/tex]
[tex]\mathbb V(Z)=\mathbb E((2+10X)^2)-\mathbb E(2+10X)^2[/tex]
[tex]\mathbb V(Z)=\mathbb E(4+40X+100X^2)-(\mathbb E(2)+\mathbb E(10X))^2[/tex]
[tex]\mathbb V(Z)=\mathbb E(4)+40\mathbb E(X)+100\mathbb E(X^2)-(\mathbb E(2)+10\mathbb E(X))^2[/tex]
[tex]\mathbb V(Z)=100\mathbb E(X^2)-40000[/tex]
[tex]\mathbb V(Z)=100\left(\mathbb E(X^2)-\mathbb E(X)^2+\mathbb E(X)^2\right)-40000[/tex]
[tex]\mathbb V(Z)=100\left(\mathbb V(X)+\mathbb E(X)^2\right)-40000[/tex]
[tex]\mathbb V(Z)=100\left(5^2+400\right)-40000[/tex]
[tex]\mathbb V(Z)=2500[/tex]
The standard deviation is the square root of the variance, so for (a) you have
[tex]\sqrt{\mathbb V(Z)}=\sqrt{2500}=50[/tex]
That should give you an idea as to how to figure out the rest.