How to reach from
$1+= [1+×]^1/t $
the power is (1/T)
to
$$APR = \frac{\ (1+EAR)^T - 1 \ }{T}$$
$$1+EAR=[1+T\times APR]^{1/T}\\ APR=\frac {(1+EAR)^T-1}T$$
and the same goes here:
$Var(aX) + Var(bY) + 2cov(aX,bY) = a^2var(X) + b^2var(Y) + 2abcov(X,Y) $
It's supposed to be mathematically simple, maybe a concept is missing for me, so I'd really appreciate if it's solved showing details as much as possible. Thanks
Beginning with $1+= (1+ \times )^\frac{1}{T}$, the crucial step is to raise both sides of the equation to the power $T$ : this gives the equivalent equation $$ (1+EAR)^T = 1+ \times $$ from which after simple algebraic cancellations you can deduce the end result.
For the second part of your question, this has nothing to do with powers, really, and everything to do with properties of the Variance and Covariance, namely,
$$\operatorname{Var}(aX) = a^2 \operatorname{Var}(X)$$ and \begin{align} \operatorname{cov}(aX, Y) = a\ \operatorname{cov}(X,Y)\\ \operatorname{cov}(X, bY) = b\ \operatorname{cov}(X,Y)\\ \end{align}
I hope this answers your questions(s). If not, I can give more details.
EDIT
From the definition of the variance,
$\operatorname{Var}(X) = \sum_s p(s) (X(s) - \operatorname{E}[X])^2$,
you get
\begin{align} \operatorname{Var}[aX] &= \sum_s p(s) (aX(s) - \operatorname{E}[aX])^2\\ &= \sum_s p(s) \left(a X(s) - a \operatorname{E}[X]\right)^2\\ &\text{(since the expected value is linear w.r.t. its variable)}\\ &= \sum_s p(s) a^2 \left(X(s) - \operatorname{E}[X]\right)^2\\ &= a^2 \operatorname{Var}[X]\\ \end{align}
and similarly for the covariance.