Given,
\[
\mu = np,
\sigma^2 = np(1-p).
\]
From \(\mu = 2\sigma^2\),
\[
np = 2 np(1-p) \implies 1 = 2(1-p) \implies 1 = 2 - 2p \implies p = \frac{1}{2}.
\]
From \(\mu + \sigma^2 = 3\),
\[
np + np(1-p) = 3 \implies np + np \times \frac{1}{2} = 3 \implies np \times \frac{3}{2} = 3 \implies np = 2.
\]
So, \(np = 2\), \(p = \frac{1}{2}\), so \(n = \frac{2}{p} = 4\).
Calculate
\[
P(X \leq 3) = \sum_{k=0}^3 \binom{4}{k} \left(\frac{1}{2}\right)^4 = 1 - P(X=4).
\]
\[
P(X=4) = \binom{4}{4} \left(\frac{1}{2}\right)^4 = \frac{1}{16}.
\]
Hence,
\[
P(X \leq 3) = 1 - \frac{1}{16} = \frac{15}{16}.
\]